Triangles
Name: _____________________
Date: _____________________
Instructions: Answer all questions. Write your answers clearly in the space provided.
If angle bisector of a triangle bisects the opposite side, then what type of triangle is it?
If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is
In a triangle ABC, BC is produced to D so that CD = AC. If ∠BAD = 111° and ∠ACB = 80°, then the measure of ∠ABC is:
In a ΔABC, AB = BC, ∠B = x° and ∠A = (2x - 20)°, Then ∠B is :
If two angles of a triangle are 21° and 38°, then the triangle is :
In ΔPQR, S and T are point on sides PR and PQ respectively such that ∠PQR = ∠PST, If PT = 5 cm, PS = 3 cm and TQ = 3 cm, then length of SR is
In a ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is :
In ΔABC, two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE. Then ΔABC and ΔDBE are :
If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then
In ΔABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.
ABC is a triangle, PQ is line segment intersecting AB is P and AC in Q and PQ || BC. The ratio of AP : BP = 3 : 5 and length of PQ is 18 cm. The length of BC is:
Which of the following is a true statement
ΔABC is similar to ΔDEF is area of ΔABC is 9 sq. cm. and area of ΔDEF is 16 sq. cm. and BC = 21 cm. Then the length of EF will be:
In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle
ABC is an isosceles triangle such that AB = AC and ∠B = 35°, AD is the median to the base BC. Then ∠BAD is
In a triangle ABC, AB = AC, ∠BAC = 40° then the external angle at B is :
In ΔABC ∠A = 90° and AD ⊥ BC where D lies on BC. If BC = 8 cm, AD = 6 cm, then arΔABC : arΔACD = ?
The angle between the external bisectors of two angles of a triangle is 60°. Then the third angle of the triangle is
In ΔABC, D and E are points on AB and AC respectively such that DE || BC and DE divides the ΔABC into two parts of equal areas. Then ratio of AD and BD is
ABC is an isosceles triangle with AB = AC, A circle through B touching AC at the middle point intersects AB at P. Then AP : AB is:
Taking any three of the line segments out of segments of length 2 cm, 3 cm, 5 cm and 6 cm, the number of triangles that can be formed is:
If the median drawn on the base of a triangle is half of its base the triangle will be
I is the incentre of ΔABC. If ∠ABC = 60°, ∠BCA = 80°, then the ∠BIC is
In a triangle ABC, incentre is O and ∠BOC = 110°, then the measure of ∠BAC is:
I is the incentre of a triangle ABC. If ∠ACB = 55°, ∠ABC = 65° then the value of ∠BIC is
For a triangle base is 6$$sqrt 3 $$ cm and two base angles are 30° and 60°. Then height of the triangle is
D is any point on side AC of ΔABC. If P, Q, X, Y are the mid-point of AB, BC, AD and DC respectively, then the ratio of PX and QY is
In ΔABC, ∠BAC = 90° and AB = $$frac{1}{2}$$ BC, Then the measure of ∠ACB is :
ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then
If in a triangle, the orthocentre lies on vertex, then the triangle is
The length of the three sides of a right angled triangle are (x - 2) cm, (x) cm and (x + 2) cm respectively. Then the value of x is
The length of the two sides forming the right angle of a right angled triangle are 6 cm and 8 cm. The length of its circum-radius is :
In a triangle ABC, ∠A = 90°, ∠C = 55°, $${AD}$$ ⊥ $${BC}$$. What is the value of ∠BAD ?
Angle between the internal bisectors of two angles of a triangle ∠B and ∠C is 120°, then ∠A is :
G is the centroid of the equilateral ΔABC. If AB = 10 cm then length of AG is ?
ABC is a right-angled triangle with AB = 6 cm and BC = 8 cm. A circle with center O has been inscribed inside ΔABC. The radius of the circle is
A point D is taken on the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then
ABC is an equilateral triangle and CD is the internal bisector of ∠C. If DC is produced to E such that AC = CE, then ∠CAE is equal to
If each angle of a triangle is less than the sum of the other two, then the triangle is
In ΔABC and ΔDEF, AB = DE and BC = EF, then one can infer that ΔABC ≅ ΔDEF, when
In triangle ABC, ∠BAC = 75°, ∠ABC = 45°, $$overline {BC} $$ is produced to D. If ∠ACD = x°, then $$frac{x}{3}$$% of 60° is
The angles of a triangle are in the ratio 2 : 3 : 7. The measure of the smallest angle is :
In ΔABC, ∠BAC = 90° and AD ⊥ BC. If BD = 3 cm and CD = 4 cm, then length of AD is :
Let ABC be an equilateral triangle and AD perpendicular to BC, then AB 2 + BC 2 + CA 2 = ?
In a ΔABC, If 2∠A = 3∠B = 6∠C, then the value of ∠B is:
In ΔABC, AD ⊥ BC and AD 2 = BD × DC. The measure of ∠BAC is :
In ΔABC, AB = BC = K, AC = $$sqrt 2 $$ k, then ΔABC is a :
In ΔABC and ΔPQR, ∠B = ∠Q, ∠C = ∠R. M is the midpoint on QR, If AB : PQ = 7 : 4, then $$frac{{{ ext{area}},left( {vartriangle ABC}
ight)}}{{{ ext{area}},left( {vartriangle PMR}
ight)}}$$ xa0 is :
In ΔABC, ∠B = 70° and ∠C = 30°, AD and AE are respectively the perpendicular on side BC and bisector of ∠A. The measure of ∠DAE is:
In ΔABC and ΔDEF, if ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠D = 60°, ∠E = 70° and ∠F = 50°, then
In ΔABC, the line parallel to BC intersect AB & AC at P & Q respectively. If AB : AP = 5 : 3, then AQ : QC is:
Given that the ratio of altitudes of two triangles is 4 : 5, ratio of their areas is 3 : 2, the ratio of their corresponding bases is :
In ΔABC, the external bisectors of the angles ∠B and ∠C meet at the point O. If ∠A = 70°, then the measure of ∠BOC is :
If I be the incentre of ΔABC and ∠B = 70° and ∠C = 50°, then the magnitude of ∠BIC is
In ΔABC, if AD ⊥ BC, then AB 2 + CD 2 is equal to
If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB = 19 cm and AC = 22 cm then the length of BC is :
ABC is a triangle in which ∠A = 90°. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP = ?
∠A of ΔABC is a right angle. AD is perpendicular on BC. If BC = 14 and BD = 5 cm, then measure of AD is:
For a triangle ABC, D, E, F are the mid - point of its sides. If ΔABC = 24 sq. units then ΔDEF is :
∠A + $$frac{1}{2}$$ ∠B + ∠C = 140°, then ∠B is :
If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60°, BC = 8 cm, then area of the triangle ABC is:
Consider the following statements :
I. Every equilateral triangle is necessarily an isosceles triangle. II. Every right-angled triangle is necessarily an isosceles triangle. III. A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle. The correct statements are:
Consider the following statements : I. Three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent. II. If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent. Of these statements :
In ΔABC, AD ⊥ BC, then
In the adjoining figure AB, EF and CD are parallel lines. Given that GE = 5 cm, GC = 10 cm and DC = 18 cm, then EF is equal to:
A triangle cannot be drawn with the following three sides
The in-radius of an equilateral triangle is of length 3 cm. Then the length of each of its medians is
The sides of a triangle are in the ratio 3 : 4 : 6. The triangle is:
In ΔABC, AD is the internal bisector of ∠A, meeting the side BC at D. If BD = 5 cm, BC = 7.5 cm, then AB : AC is
If the circumradius of an equilateral triangle be 10 cm, then the measure of its in-radius is
I is the incentre of ΔABC, ∠ABC = 60° and ∠ACB = 50°, Then ∠BIC is
If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -
The circumcentre of a triangle ABC is O. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠OAC is
If the incentre of an equilateral triangle lies inside the triangle and its radius in 3 cm, then the side of the equilateral triangle is
ΔABC be a right-angled triangle where ∠A = 90° and AD ⊥ BC. If ar (ΔABC) = 40 cm 2 , ar (ΔACD) = 10 cm 2 and AC = 9 cm, then the length of BC is
The orthocentre of a right angled triangle lies
O is the incentre of ΔABC and ∠A = 30°, then ∠BOC is
If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5 cm then AB is:
In a triangle ABC, ∠BAC = 90° and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm then the length of BC is:
Let O be the in-centre of a triangle ABC and D be a point on the side BC of ΔABC, such that OD ⊥ BC. If ∠BOD = 15°, then ∠ABC = ?
If the circumcentre of a triangle lies outside it, then the triangle is
An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75° = ?
In a triangle ABC, the side BC is extended up to D such that CD = AC. If ∠BAD = 109° and ∠ACB = 72° then the value of ∠ABC is
The equidistant point from the vertices of a triangle is called its:
Let ABC be an equilateral triangle and AX, BY, CZ be the altitudes. Then the right statement out of the four given responses is
ABC is an isosceles triangle with AB = AC. The side BA is produced to D such that AB = AD. If ∠ABC = 30°, then ∠BCD is equal to
The sum of three altitudes of a triangle is
In ΔABC, ∠B = 60° and ∠C = 40°. If AD and AE be respectively the internal bisector of ∠A and perpendicular on BC, then the measure of ∠DAE is
In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angle must be
In a ΔABC ∠A : ∠B : ∠C = 2 : 3 : 4. A line CD drawn || to AB, then the ∠ACD is :
BL and CM are medians of ΔABC right-angled at A and BC = 5 cm. If BL = $$frac{{3sqrt 5 }}{2}$$ cm, then the length of CM is
ABC is an equilateral triangle. Points D, E and F are taken as the mid-point on sides AB, BC, AC respectively, so that AD = BE = CF. Then AE, BF, CD enclosed a triangle which is:
The side BC of a triangle ABC is proceed to D. If ∠ACD = 112° and ∠B = $$frac{3}{4}$$ ∠A, then the measure of ∠B is:
ΔABC is similar to ΔDEF. If the sides of ΔABC, that is AB, BC and CA, are 3, 4 and 5 cms respectively, what would be the perimeter of the ΔDEF, if the side DE measures 12 cms ?
In a ΔPQR, ∠Q = 55° and ∠R = 35°. Find the ratio of angles subtended by side QR on circumcenter, incenter and orthocenter of the triangle.
In ΔPQR, straight line parallel to the base QR cuts PQ at X and PR at Y. If PX : XQ = 5 : 6, then XY : QR will be
In an isosceles triangle ΔABC, AB = AC and ∠A = 80°. The bisector of ∠B and ∠C meet at D. The ∠BDC is equal to.
Let ΔABC and ΔABD be on the same base AB and between the same parallels AB and CD. Then the relation between areas of triangles ABC and ABD will be
G is the centroid of ΔABC. If AB = BC = AC, then measure of ∠BGC is:
In a triangle ABC, if ∠A + ∠C = 140° and ∠A + 3∠B = 180°, then ∠A is equal to:
If the measures of the sides of triangle are (x 2 - 1), (x 2 + 1) and 2x cm, then the triangle would be :
In ΔABC, ∠C is an obtuse angle. The bisectors of the exterior angles at A and B meet BC and AC produced at D and E respectively. If AB = AD = BE, then ∠ACB = ?
Let ABC be an equilateral triangle and AX, BY, CZ be the altitude. Then the right statement out of the four give responses is :
In ΔABC, DE || AC, D and E are two points on AB and CB respectively. If AB = 10 cm and AD = 4 cm, then BE : CE is
If the three angles of a triangle are: $${left(x + 15
ight)^ circ },$$ xa0 $${left({frac{{6x}}{5} + 6}
ight)^ circ }$$ xa0and $${left({frac{{2x}}{3} + 30}
ight)^ circ }$$ xa0 then the triangle is:
If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $$frac{{AD}}{{BD}}$$ = $$frac{3}{5}$$. If AC = 4 cm, then AE is
In a ΔABC, ∠A + ∠B = 118°, ∠A + ∠C = 96°. Find the value of ∠A.
For a triangle ABC, D and E are two points on AB and AC such that AD = $$frac{1}{4}$$ AB, AE = $$frac{1}{4}$$ AC. If BC = 12 cm, then DE is :
In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD, then DE : BC is
ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°, then the value of ∠GAB is :
If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is:
In the following figure which of the following statements is true?
In triangle PQR length of the side QR is less than twice the length of the side PQ by 2 cm. Length of the side PR exceeds the length of the side PQ by 10 cm. The perimeter is 40 cm. The length of the smallest side of the triangle PQR is :
AB and CD bisect each other at O. If AD = 6 cm. Then BC is :
In a triangle ABC,∠xa0A = 90°, AL is drawn perpendicular to BC, Then ∠BAL is equal to:
Consider the triangle shown in the figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC ?
In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then length of AD is :
The point of intersection of the altitudes of a triangle is called its:
In ΔPQR, PS is the bisector of ∠P and PT ⊥ OR, then ∠TPS is equal to:
Two right angled triangles are congruent if : I. The hypotenuse of one triangle is equal to the hypotenuse of the other. II. A side for one triangle is equal to the corresponding side of the other. III. Sides of the triangles are equal. IV.xa0An angle of the triangle are equal. Of these statements, the correct ones are combination of:
In a triangle, if three altitudes are equal, then the triangle is
In an isosceles triangle, if the unequal angle is twice the sum of the equal angles, then each equal angle is
If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is :
In a right-angle ΔABC, ∠ABC = 90°, AB = 5 cm and BC = 12 cm. The radius of the circumcircle of the triangle ABC is
In triangle PQR, points A, B and C are taken on PQ, PR and QR respectively such that QC = AC and CR = CB. If ∠QPR = 40°, then ∠ACB is equal to:
If ABC is an equilateral triangle and D is a point of BC such that AD ⊥ BC, then
ΔABC is an isosceles triangle and $$overline {AB} $$xa0 = $$overline {AC} $$xa0 = 2a unit, $$overline {BC} $$xa0 = a unit. Draw $$overline {AD} $$ ⊥ $$overline {BC} $$ , and find the length of $$overline {AD} $$
ABC is a triangle. The bisectors of the internal angle ∠B and external angle ∠C intersect at D. If ∠BDC = 50°, then ∠A is
AD is the median of a triangle ABC and O is the centroid such that AO = 10 cm. The length of OD (in cm) is
The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ∠PSR is
In a right angled triangle ΔDEF, if the length of the hypotenuse EF is 12 cm, then the length of the median DX is:
In an equilateral triangle ABC, G is the centroid. Each side of the triangle is 6 cm. The length of AG is:
In case of an acute angled triangle, its orthocenter lies:
If the measure of the angles of a triangle are in the ratio 1 : 2 : 3 and if the length of the smallest side of the triangle is 10 cm, then the length of the longest side is:
In a triangle ABC, ∠A = 70°, ∠B = 80° and D is the incenter of ΔABC, ∠ACB = 2x° and ∠BDC = y°. The values of x and y, respectively are:
In ΔABC, AC = BC and ∠ABC = 50°, the side BC is produced to D so that BC = CD then the value of ∠BAD is:
An equilateral triangle of side 6 cm is inscribed in a circle. Then radius of the circle is:
If ΔPQR and ΔLMN are similar and 3PQ = LM and MN = 9 cm, then QR is equal to:
In a ΔABC, ∠A + ∠B = 75° and ∠B + ∠C = 140°, then ∠B is:
ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?
Length of the sides of a triangle are a, b and c respectively. If a 2 + b 2 + c 2 = ab + bc + ca then the triangle is:
PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is:
Incenter of ΔABC is I. ∠ABC = 90° and ∠ACB = 70°. ∠BIC is:
If in ΔABC, DE || BC, AB = 7.5 cm BD = 6 cm and DE = 2 cm then the length of BC in cm is:
Which of the set of three sides can't form a triangle?
The orthocenter of a triangle is the point where?
Possible length of the sides of a triangle are:
The centroid of a triangle is G. If area of ΔABC = 72 sq. unit, then the area of ΔBGC is?
Three sides of a triangle are 5 cm, 9 cm and x cm. The minimum integral value of x is:
ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12$$sqrt 5 $$ and BC = 24 cm then radius of circle is:
In a ΔABC, BC is extended upto D
∠ACD = 120°, ∠B = $$frac{1}{2}$$ ∠A, then ∠A is:
In ΔABC, ∠B = 60° and ∠C = 40°
AD and AE are respectively the bisector of ∠A and perpendicular on BC. The measure of ∠EAD is:
In a right angled ΔABC, ∠ABC = 90°, AB = 3, BC = 4, CA = 5
BN is perpendicular to AC, AN : NC is
Answer Key
& frac{{PR}}{{PT}} = frac{{PQ}}{{PS}} cr
& frac{{PR}}{5} = frac{8}{3} cr
& PR = frac{{40}}{3} cr
& herefore SR = PR - PS cr
& SR = frac{{40}}{3} - 3 cr
& SR = frac{{40 - 9}}{3} cr
& SR = frac{{31}}{3},{ ext{cm}} cr} $$
& frac{{AP}}{{AB}} = frac{{PQ}}{{BC}} cr
& frac{3}{8} = frac{{PQ}}{{BC}} cr
& frac{3}{8} = frac{{18}}{{BC}} cr
& BC = 48,{ ext{cm}} cr} $$
& herefore frac{{AB}}{{DE}} = frac{{BC}}{{EF}} = frac{{sqrt 9 }}{{sqrt {16} }} cr
& = frac{{2.1}}{{EF}} = frac{3}{4} cr
& EF = 2.8,{ ext{cm}} cr} $$
& herefore frac{{{ ext{area}},{ ext{of}},Delta ABC}}{{{ ext{area}},{ ext{of}},Delta ACD}} = frac{{B{C^2}}}{{A{D^2}}} cr
& frac{{{ ext{area}},{ ext{of}},Delta ABC}}{{{ ext{area}},{ ext{of}},Delta ACD}} = frac{{{8^2}}}{{{6^2}}} cr
& frac{{{ ext{area}},{ ext{of}},Delta ABC}}{{{ ext{area}},{ ext{of}},Delta ACD}} = frac{{64}}{{36}} cr
& frac{{{ ext{area}},{ ext{of}},Delta ABC}}{{{ ext{area}},{ ext{of}},Delta ACD}} = frac{{16}}{9} cr} $$ area ΔABC : area ΔACD = 16 : 9
& frac{{{ ext{ar}},Delta ADE}}{{{ ext{ar}},Delta ABC}} = frac{{A{D^2}}}{{A{B^2}}} cr
& frac{1}{2} = {left( {frac{{AD}}{{AB}}}
ight)^2} cr
& frac{1}{{sqrt 2 }} = frac{{AD}}{{AB}} cr
& herefore frac{{AD}}{{DB}} = frac{1}{{sqrt 2 - 1}} cr
& left( { herefore DB = AB - AD = sqrt 2 - 1}
ight) cr
& { ext{So,}},AD:BD = 1:sqrt 2 - 1 cr} $$
& AP = frac{x}{2} cr
& frac{{AP}}{{AB}} = frac{x}{{2 imes 2x}} = frac{1}{4} cr
& frac{{AP}}{{AB}} = frac{1}{4} cr
& AP:AB = 1:4 cr} $$
& frac{{AB}}{{BC}} = frac{1}{2},,,,,,,,,,,,,,,,frac{P}{H} = frac{1}{2} cr
& sin heta = frac{P}{H} = frac{1}{2} cr
& sin ,{30^ circ } = frac{1}{2} cr
& herefore heta = angle ACB = {30^ circ } cr} $$
& {left( {frac{{ab}}{p}}
ight)^2} = {a^2} + {b^2} cr
& Rightarrow frac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} cr
& Rightarrow frac{1}{{{p^2}}} = frac{{{a^2}}}{{{a^2}{b^2}}} + frac{{{b^2}}}{{{a^2}{b^2}}} cr
& Rightarrow frac{1}{{{p^2}}} = frac{1}{{{a^2}}} + frac{1}{{{b^2}}} cr} $$ Alternate : From figure, $$eqalign{
& P = frac{{ab}}{c} cr
& P = frac{{ab}}{{sqrt {{a^2} + {b^2}} }},left( {x08ecause {a^2} + {b^2} = {c^2}}
ight) cr
& {P^2} = frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}} cr
& frac{1}{{{p^2}}} = frac{1}{{{a^2}}} + frac{1}{{{b^2}}} cr} $$
& left( {x - 2}
ight),,,,,,,,,,,,,x,,,,,,,,,,,,,left( {x + 2}
ight) cr
& ,,,,,,, downarrow ,,,,,,,,,,,,,,,,,,,,,,, downarrow ,,,,,,,,,,,,,,,,,,,,,, downarrow cr
& ,,,,,,,6,,,,,,,,,,,,,,,,,,,,,,,8,,,,,,,,,,,,,,,,,,,,10 cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, downarrow cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{Triplet}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{x}} = { ext{8}} cr} $$
& = frac{{sqrt 3 }}{2} imes 10 = 5sqrt 3 cr
& herefore 3,{ ext{units}} = 5sqrt 3 cr
& ,,,,,,1,{ ext{unit}} = frac{{5sqrt 3 }}{3} cr
& ,,,,,,2,{ ext{units}} = frac{{5sqrt 3 }}{3} imes 2 cr
& ,,,,,,2,{ ext{units}} = frac{{10sqrt 3 }}{3} cr
& herefore { ext{AG}} = frac{{10sqrt 3 }}{3},cm cr} $$
& = frac{{a + b - c}}{2} cr
& = frac{{8 + 6 - 10}}{2} cr
& = frac{4}{2} cr
& = 2,cm cr} $$
ight)$$
& 2angle A = 3angle B cr
& frac{{angle A}}{{angle B}} = frac{3}{2},,,,,,,,,,,,,3angle B = 6angle C cr
& frac{{angle B}}{{angle C}} = frac{6}{3} = frac{2}{1} cr} $$ To make angle ∠B same ∴ ∠A xa0 xa0 : xa0 xa0 ∠B xa0 xa0 : xa0 xa0 ∠C 3 xa0 xa0 : xa0 xa0xa0 xa0 2 xa0 xa0 : xa0 xa0 1 As we know that ∠A + ∠B + ∠C = 180° 3x + 2x + x = 180° x = 30° ∠B = 2x ∠B = 60°
& frac{{{ ext{area}}{mkern 1mu} left( {vartriangle ABC}
ight)}}{{{ ext{area}}{mkern 1mu} left( {vartriangle PMR}
ight)}} cr
& = frac{{{{left( 7
ight)}^2}}}{{frac{1}{2} imes {{left( 4
ight)}^2}}} cr
& = frac{{49}}{8} cr} $$
& frac{{AP}}{{PB}} = frac{{AQ}}{{QC}} cr
& frac{{AQ}}{{QC}} = frac{3}{2} cr
& AQ:QC = 3:2 cr} $$
& frac{{{ ext{area}},{ ext{of}},{ ext{triangle}},1}}{{{ ext{area}},{ ext{of}},{ ext{triangle}},2}} = frac{3}{2} cr
& Rightarrow frac{{frac{1}{2} imes {B_1} imes 4}}{{frac{1}{2} imes {B_2} imes 5}} = frac{3}{2} cr
& Rightarrow frac{{{B_1}}}{{{B_2}}} = frac{3}{2} imes frac{5}{4} cr
& Rightarrow frac{{{B_1}}}{{{B_2}}} = frac{{15}}{8} cr
& herefore {B_1}:{B_2} = 15:8 cr} $$
& frac{{{ ext{GE}}}}{{{ ext{CG}}}} = frac{{{ ext{EF}}}}{{{ ext{CD}}}} cr
& { ext{or, }}frac{5}{{10}} = frac{{{ ext{EF}}}}{{18}} cr
& { ext{or,}},{ ext{EF}} = 9,{ ext{cm}} cr} $$
& frac{{AB}}{{AC}} = frac{{BD}}{{DC}} cr
& frac{{AB}}{{AC}} = frac{5}{{2.5}} cr
& frac{{AB}}{{AC}} = frac{2}{1} cr
& herefore AB:AC = 2:1 cr} $$
& frac{{40}}{{10}} = frac{{B{C^2}}}{{{{left( 9
ight)}^2}}} cr
& frac{{40}}{{10}} imes 81 = B{C^2} cr
& BC = 18,{ ext{cm}} cr} $$
& AB = sqrt {B{D^2} + A{D^2}} cr
& AB = sqrt {{4^2} + {6^2}} cr
& AB = sqrt {52} ,cm cr} $$ ΔBAC ∼ ΔBDA $$eqalign{
& herefore frac{{BC}}{{AB}} = frac{{AB}}{{BD}} cr
& herefore frac{{BC}}{{sqrt {52} }} = frac{{sqrt {52} }}{4} cr
& BC = frac{{52}}{4} cr
& BC = 13,cm cr} $$ Alternate : $$eqalign{
& A{B^2} = BD.BC cr
& {left( {sqrt {B{D^2} + A{D^2}} }
ight)^2} = BD.BC cr
& {left( {sqrt {{4^2} + {6^2}} }
ight)^2} = 4.BC cr
& frac{{52}}{4} = BC, cr
& herefore BC = 13,cm cr} $$
& sin {75^ circ } = frac{{AP}}{{AD}} cr
& sin {75^ circ } = frac{a}{{2b}} imes sqrt 3 cr
& sin {75^ circ } = frac{{sqrt 3 a}}{{2b}} cr} $$
& angle DAE = frac{{angle B - angle C}}{2} cr
& ,,,,,,,,,,,,,,,,,,,, = frac{{{{60}^ circ } - {{40}^ circ }}}{2} cr
& ,,,,,,,,,,,,,,,,,,,, = {10^ circ } cr} $$
ight)^2}$$ xa0+ 4CM 2 = 5BC 2 ⇒ 45 + 4CM 2 = 125 ⇒ CM 2 = $$frac{{125 - 45}}{4}$$ ⇒ CM 2 = 20 ⇒ CM = $$2sqrt 5 $$xa0 cm
& {x^ circ } + frac{3}{4}{x^ circ } = {112^ circ } cr
& frac{{7{x^ circ }}}{4} = {112^ circ } cr
& {x^ circ } = {64^ circ } cr
& { ext{Hence,}} cr
& angle B = frac{3}{4} imes {64^ circ } cr
& angle B = {48^ circ } cr} $$
& frac{{{ ext{Perimeter}},{ ext{of}},vartriangle ABC}}{{{ ext{Perimeter}},{ ext{of}},vartriangle DEF}} = frac{{AB}}{{DE}} cr
& frac{{12}}{{{ ext{Perimeter}},{ ext{of}},vartriangle DEF}} = frac{3}{{12}} cr} $$ Perimeter of ΔDEF = 48 cms
& frac{{PX}}{{PQ}} = frac{{XY}}{{QR}} cr
& frac{5}{{left( {5 + 6}
ight)}} = frac{{XY}}{{QR}} cr
& XY:QR = 5:11 cr} $$
& Rightarrow angle CAD = frac{{180 - x}}{2} = 90 - frac{x}{2} cr
& { ext{and}} cr
& Rightarrow angle EBC = frac{{180 - y}}{2} = 90 - frac{y}{2} cr
& { ext{also}},angle AEB = angle EAB = x cr} $$ (∵ AB = EB ⇒ ABE is an isosceles triangle) and ∠ADB = ∠ABD = y (∵ AB = AD ⇒ ADB is an isosceles triangle) In ΔAEB, ∠AEB + ∠ABE + ∠BAE = 180° x + x + y + 90 - $$frac{y}{2}$$ = 180° ⇒ 4x + y = 180° Similarly in ΔADB 4y + x = 180° ⇒ 4y + x + 4x + y = 180 + 180 ⇒ 5x + 5y = 360° ⇒ x + y = 72° In triangle ABC, ∠ACB + x + y = 180° ⇒ ∠ACB = 180 - 72 ⇒ ∠ACB = 108°
& herefore frac{{BD}}{{AD}} = frac{{BE}}{{CE}} cr
& ,,,,,,frac{{BE}}{{CE}} = frac{6}{4} cr
& ,,,,,,frac{{BE}}{{CE}} = frac{3}{2} cr
& ,,,,,,BE:CE = 3:2 cr} $$
ight) + {left( {frac{{6x}}{5} + 6}
ight)^ circ } + $$ xa0xa0 xa0 $${left( {frac{{2x}}{3} + 30}
ight)^ circ } = $$ xa0xa0 $${180^ circ }$$ $$,,,,,,,,,,,,left{ {angle A + angle B + angle C = {{180}^ circ }}
ight}$$ $$ Rightarrow x + frac{{6x}}{5} + frac{{2x}}{3} = $$ xa0 xa0 $${180^ circ } - left(15 + 6 + 30
ight)$$ $$eqalign{
& Rightarrow frac{{15x + 18x + 10x}}{{15}} = 180 - 51 cr
& Rightarrow 43x = 129 imes 15 cr
& Rightarrow x = {45^ circ } cr
& Rightarrow { ext{each}},{ ext{angle}} cr
& Rightarrow {left( {x + 15}
ight)^ circ } = 45 + 15 = {60^ circ } cr
& Rightarrow {left( {frac{{6x}}{5} + 6}
ight)^ circ } = {60^ circ } cr
& Rightarrow {left( {frac{{2x}}{3} + 30}
ight)^ circ } = {60^ circ } cr} $$ ∵ All three angles are equal 60° ⇒ Triangle will be equilateral triangle
& frac{{AD}}{{AB}} = frac{{AE}}{{AC}} = frac{{DE}}{{BC}} cr
& frac{3}{8} = frac{{AE}}{4} cr
& AE = frac{3}{2} = 1.5,{ ext{cm}} cr} $$
& frac{{AD}}{{AB}} = frac{{AE}}{{AC}} = frac{{DE}}{{BC}} cr
& frac{{AD}}{{AB}} = frac{{DE}}{{BC}} cr
& Rightarrow ,frac{1}{4} = frac{{DE}}{{12}} cr
& Rightarrow DE = 3,{ ext{cm}} cr} $$
& frac{{AD}}{{AB}} = frac{{DE}}{{BC}} = frac{{AE}}{{AC}} cr
& frac{{DE}}{{BC}} = frac{1}{2} cr
& herefore DE:BC = 1:2 cr} $$
PQ + QR + Rp = 40 Putting the value of PQ and QR from equation (1) and (2), PQ + 2PQ - 2 + PQ + 10 = 40 4PQ = 32 PQ = 8 cm which is the smallest side of the triangle.
& frac{{{ ext{Perimeter of Delta ADC}}}}{{{ ext{Perimeter of Delta BDC}}}} cr
& = frac{{6 + 7 + 8}}{{9 + 6 + 12}} cr
& = frac{{21}}{{27}} cr
& = frac{7}{9} cr} $$
∠R = 90° -∠2 - ∠3
So,
∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3)
∠Q - ∠R = ∠2 + ∠3 - ∠1
∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1 [using equation 1] ∠Q - ∠R = 2∠2
$$frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS
& herefore frac{{AB}}{{BD}} = frac{2}{1} cr
& AB:BD = 2:1 cr} $$
ight)^2}$$ + AD 2 4a 2 = $$frac{{{a^2}}}{4}$$ + AD 2 AD 2 = 4a 2 - $$frac{{{a^2}}}{4}$$ AD = $$frac{{sqrt {15} }}{2}$$ a units
& = frac{{EF}}{2} cr
& = frac{{12}}{2} cr
& = 6,{ ext{cm}} cr} $$
& AG:GD = 2:1 cr
& AD = frac{{sqrt 3 }}{2}a cr
& AD = frac{{sqrt 3 }}{2} imes 6 cr
& AD = 3sqrt 3 cr
& 3 o 3sqrt 3 cr
& 1 o sqrt 3 cr
& AG = 2,{ ext{unit}} = 2sqrt 3 ,{ ext{cm}} cr} $$
& R = frac{a}{{sqrt 3 }} cr
& R = frac{6}{{sqrt 3 }} cr
& R = 2sqrt 3 ,{ ext{cm}} cr} $$
& frac{{PQ}}{{LM}} = frac{{QR}}{{MN}} = frac{{PR}}{{LN}} cr
& left( {vartriangle PQR,{ ext{and}},vartriangle LMN,{ ext{are}},{ ext{similar}}}
ight) cr
& frac{{PQ}}{{LM}} = frac{{QR}}{{MN}} cr
& frac{1}{3} = frac{{QR}}{9} cr
& QR = 3,{ ext{cm}} cr} $$
& vartriangle ADE simeq vartriangle ABC cr
& frac{{AD}}{{AB}} = frac{{DE}}{{BC}} cr
& frac{{1.5}}{{7.5}} = frac{2}{{BC}} cr
& BC = 10,{ ext{cm}} cr} $$
& {R_2} = frac{{abc}}{{4vartriangle }} cr
& vartriangle = sqrt {Sleft( {s - a}
ight)left( {s - b}
ight)left( {s - c}
ight)} cr} $$ $$ = sqrt {12left( {sqrt 5 + 1}
ight)left( {12}
ight) imes 12 imes 12left( {sqrt 5 - 1}
ight)} $$ where a = 12$$sqrt 5 $$ , b = 12$$sqrt 5 $$xa0 & c = 24 $$eqalign{
& S = frac{{a + b + c}}{2} cr
& S = frac{{24sqrt 5 + 24}}{2} cr
& S = 12left( {sqrt 5 + 1}
ight) cr
& {R_2} = frac{{12sqrt 5 imes 12sqrt 5 imes 24}}{{4 imes 12 imes 12 imes 2}} cr
& {R_2} = frac{{30}}{2} cr
& {R_2} = 15,{ ext{cm}} cr} $$
& angle EAD = frac{{angle B - angle C}}{2} cr
& angle EAD = frac{{60 - 40}}{2} cr
& angle EAD = {10^ circ } cr} $$
& frac{{AB}}{{BN}} = frac{{AC}}{{BC}} cr
& BN = frac{{AB imes BC}}{{AC}} = frac{{3 imes 4}}{5} = 2.4 cr
& frac{{BC}}{{NC}} = frac{{AB}}{{NB}} cr
& frac{4}{{NC}} = frac{3}{{2.4}} cr
& NC = 3.2 cr
& frac{{AB}}{{AN}} = frac{{BC}}{{NB}},,,,,,,,,,,,,frac{3}{{AN}} = frac{4}{{2.4}} cr
& AN = 1.8 cr
& frac{{AN}}{{NC}} = frac{{1.8}}{{3.2}} cr
& frac{{AN}}{{NC}} = frac{9}{{16}} cr
& herefore AN:NC = 9:16 cr} $$