Plane Geometry
- In ∆ PQR, 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
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On the basis of question we draw a figure of triangle PQR ,
XY || QR
∠PXY = ∠PQR
∠PYX = ∠PRQ
By AA–similarity,
∆PXY ~ ∆PQR∴ PQ = QR PX XY ∵ PX = 5 QX 6 ⇒ XQ = 6 PX 5 ⇒ XQ + PX = 6 + 5 = 5 : 11 PX 5 ⇒ PQ = 11 PX 5
Correct Option: A
On the basis of question we draw a figure of triangle PQR ,
XY || QR
∠PXY = ∠PQR
∠PYX = ∠PRQ
By AA–similarity,
∆PXY ~ ∆PQR∴ PQ = QR PX XY ∵ PX = 5 QX 6 ⇒ XQ = 6 PX 5 ⇒ XQ + PX = 6 + 5 = 5 : 11 PX 5 ⇒ PQ = 11 PX 5 ⇒ PX = XY = 5 PQ QR 11
- The mid points of AB and AC of the ∆ ABC are P and Q respectively. If PQ = 6 cm., then the side BC is
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As per the given in question , we draw a figure triangle ABC and P and Q are the mid points of AB and AC ,
The straight line joining the mid– points of the two sides of a triangle is parallel to the third side and half of it.Correct Option: B
As per the given in question , we draw a figure triangle ABC and P and Q are the mid points of AB and AC ,
The straight line joining the mid– points of the two sides of a triangle is parallel to the third side and half of it.
Here , PQ = 6 cm.
∴ BC = 2 × PQ = 2 × 6 = 12 cm.
- The area of two similar Δs are 121 cm2 and 81 cm2 respectively. What is the ratio of their corresponding heights (altitudes):
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In given Δ ABC and Δ DEF ,
ar(∆ABC) = 121 cm2 and ar(∆DEF) = 81 cm2
We know that ,ar(∆ABC) = AM2 ar(∆DEF) DN2 
Correct Option: A
In given Δ ABC and Δ DEF ,
ar(∆ABC) = 121 cm2 and ar(∆DEF) = 81 cm2
We know that ,ar(∆ABC) = AM2 ar(∆DEF) DN2 121 = AM2 81 DN2 ∴ AM = 11 DN 9
Thus , AM : DN = 11 : 9 .
- In Δ ABC, the median BE intersects AC at E, if BG = 6 cm, where G is the centro-id, then BE is equal to :
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We know that the centroid of a Δ divides each median in the ratio of 2 : 1
Given - BG = 6 cm∴ BG : BE = 3 : 2⇒ BE = 3 BG 2 
Correct Option: B
We know that the centroid of a Δ divides each median in the ratio of 2 : 1
Given - BG = 6 cm∴BG : BE = 3 : 2⇒ BE = 3 BG 2 ⇒ BE = 3 × 6 = 9 cm. 2 
- If the sides of a triangle are produced then the sum of the exterior angles i.e., ∠ a + ∠ b + ∠ c is equal to :
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From above figure , we have
Since every exterior angle is equal to sum of interior opposite angles,
So, ∠ a = A + B, ∠ b = B + C and ∠ c = A + C
We know that , A + B + C = 180°
∴ ∠a +∠b + ∠c = 2(A + B + C) = 2 × 180° = 360°.Correct Option: C
From above figure , we have
Since every exterior angle is equal to sum of interior opposite angles,
So, ∠ a = A + B, ∠ b = B + C and ∠ c = A + C
We know that , A + B + C = 180°
∴ ∠a +∠b + ∠c = 2(A + B + C) = 2 × 180° = 360°.
