Geometry - Study Mode

[#141] ΔPQR is an isosceles triangle and PQ = PR = 2a unit, QR = a unit. Draw PX ⊥ QR, and find the length of PX.
Correct Answer

(D) $$frac{{sqrt {15} a}}{2}$$

Explanation

Solution: $$eqalign{
& { ext{Given that,}} cr
& PQ = PR = 2a, cr
& QR = a, cr
& QX = frac{a}{2} cr
& Rightarrow {left( {2a}
ight)^2} = {left( {PX}
ight)^2} + {left( {frac{a}{2}}
ight)^2} cr
& Rightarrow 4{a^2} - frac{{{a^2}}}{4} = {left( {PX}
ight)^2} cr
& Rightarrow {left( {PX}
ight)^2} = frac{{15{a^2}}}{4} cr
& Rightarrow PX = frac{{sqrt {15} a}}{2}{ ext{ Answer}} cr} $$

[#142] The sides BA and DE of a regular pentagon are produced to meet at F. What is the measure of ∠EFA?
Correct Answer

(B) 36°

Explanation

Solution: Interior angle of a regular polygon $$eqalign{
& = frac{{left( {n - 2}
ight) imes {{180}^ circ }}}{n} cr
& = frac{{left( {5 - 2}
ight) imes {{180}^ circ }}}{5} cr
& = {108^ circ } cr} $$ ∠DEA = ∠BAE = 108° ∠AEF = ∠EAF = 72° In ΔEAF, ∠E + ∠A + ∠F = 180° 72 + 72 + ∠EFA = 180° ∠EFA = 36°

[#143] Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 24 cm, then what is the distance between their centres?
Correct Answer

(D) 14 cm

Explanation

Solution: In ΔADO OD = $$sqrt {{{left( {15}
ight)}^2} - {{left( {12}
ight)}^2}} $$ xa0 xa0= 9 cm In ΔADC DC = $$sqrt {{{left( {13}
ight)}^2} - {{left( {12}
ight)}^2}} $$ xa0 xa0= 5 cm or use Triplet 5, 12, 13 and 9, 12, 15 ∴ OC = DC + OD = 5 + 9 = 14 cm

[#144] In the given figure, SX is tangent. SX = OX = OR. If QX = 3 cm and PQ = 9 cm. Then what is the value (in cm) of OS ?
Correct Answer

(D) 3

Explanation

Solution: In the given circle, SX is a tangent SX 2 = (XQ) × (PX) SX 2 = 3 × [PQ+ QX] SX 2 = 3 × [3 + 9] SX 2 = 3 × 12 SX 2 = 36 SX = 6 cm RO = 6 = OX = SX ∴ OQ = OX - QX and PO = PQ - OQ OQ = 6 - 3 = 3 cm and PO = 9 - 3 = 6 cm PQ and RS Intersects at O, ∴ PO × OQ = RO × OS ⇒ 6 × 3 = 6 × OS ⇒ OS = 3 cm

[#145] In a circle of radius 10 cm, with centre O, PQ and PR are two chords each of length 12 cm. PO intersects chord QR at the points S. The length of OS is:
Correct Answer

(A) 2.8 cm

Explanation

Solution: QS 2 = 12 2 - (10 - x) 2 = 10 2 - x 2 12 2 - 10 2 = (10 - x) 2 - x 2 22 × 2 = 10(10 - 2x) 4.4 = 10 - 2x 2x = 5.6 x = 2.8 cm