Section (D) Long Answer Type 20 x 5M = 10 Marks8. AB and CD are respec

1.	Section (D) Long Answer Type 20 x 5M = 10 Marks8. AB and CD are respectively the smallest and longest sides of quadrilateralABCD. Show that ZA > C and B> 2D9. The side QR of a triangle PQR is produced to a point s. If the bisector of< PQR and Z PRS meet at point T, QPR (See given Figure), then prove thatAngle QTR=½Angle QPR
Answer» Answer: What would you like to ask? MATHS avatar Asked on DECEMBER 26, 2019 by Kanishka Bhul AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). SHOW that ∠A>∠C and ∠B>∠D question Share Study later ANSWER Let ABCD be a quadrilateral such that AB is it's smallest side and CD is it largest side. Now join AC and BD In ΔABC ,we have BC>AB ⇒∠8>∠3 ...(1) {∵ ANGLE opposite to longer side is greater.} Since CD is the longest side of quadrilateral ABCD In ΔACD, we have CD>AD ⇒∠7>∠4 ...(2) Adding (1) and(2) we get ∠8+∠7>∠3+∠4 ⇒∠A>∠C Now in ΔABD, we have AD>AB {∵AB is the shortest side} ⇒∠1>∠6 ...(3) In ΔBCD,we have CD>BC ⇒∠2>∠5 ....(4) Adding (3) and (4) ∠1+∠2>∠5+∠6 ⇒∠B>∠D Hence proved.

Section (D) Long Answer Type 20 x 5M = 10 Marks8. AB and CD are respectively the smallest and longest sides of quadrilateralABCD. Show that ZA > C and B> 2D9. The side QR of a triangle PQR is produced to a point s. If the bisector of< PQR and Z PRS meet at point T, QPR (See given Figure), then prove thatAngle QTR=½Angle QPR

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Asked on DECEMBER 26, 2019 by

Kanishka Bhul

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). SHOW that ∠A>∠C and ∠B>∠D

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Study later

ANSWER

Let ABCD be a quadrilateral such that AB is it's smallest side and CD is it largest side.

Now join AC and BD

In ΔABC ,we have

BC>AB

⇒∠8>∠3 ...(1) {∵ ANGLE opposite to longer side is greater.}

Since CD is the longest side of quadrilateral ABCD

In ΔACD, we have

CD>AD

⇒∠7>∠4 ...(2)

Adding (1) and(2) we get

∠8+∠7>∠3+∠4

⇒∠A>∠C

Now in ΔABD, we have

AD>AB {∵AB is the shortest side}

⇒∠1>∠6 ...(3)

In ΔBCD,we have

CD>BC

⇒∠2>∠5 ....(4)

Adding (3) and (4)

∠1+∠2>∠5+∠6

⇒∠B>∠D

Hence proved.