Question

B. Is it possible to have a triangle with the sides 6 cm , 3 cm, 2cm ?
C. PQR is a triangle, right angled at P. If PQ = 7 cm , PR = 24 cm , find QR.
D. Diagonals of a rhombus measure 6 cm and 8 cm. Find its perimeter.​

Answer 1

Answer:

C) Given: PQ=10cm, PR=24cm

C) Given: PQ=10cm, PR=24cmLet QR be x cm.

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24)

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676⇒x=

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676⇒x= 676

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676⇒x= 676

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676⇒x= 676 =26cm

C) Given: PQ=10cm, PR=24cmLet QR be x cm.In right angled triangle QPR,(Hypotenuse) 2 =(Base) 2 +(Perpendicular) 2 [By Pythagoras theorem]⇒(QR) 2 =(PQ) 2 +(PR) 2 ⇒x 2 =(10) 2 +(24) 2 ⇒x 2 =100+576=676⇒x= 676 =26cmThus, the length of QR is 26cm.

Answer 2

Question :

Is it possible to have a triangle with the sides 6 cm , 3 cm, 2cm ?

Answer:

We know that to construct any triangle the sum of two sides should be greater than third .

$2 + 3 = 5 \\ 5 < 6$

Here, Sum of two sides is not greater than third one. So we cannot construct the triangle.

Question :

PQR is a triangle, right angled at P. If PQ = 7 cm , PR = 24 cm , find QR.

Answer:

Given :

PQ = 7cm

PR = 24 cm

QR = ?

According to Pythagoras theorem :

$QR ^{2} = PQ ^{2} + PR ^{2} \\ {7}^{2} + 24 ^{2} = QR^{2} \\ = 49 + 576 \\ = \sqrt 625 = 25$

Hence, QR =25

Question :

Diagonals of a rhombus measure 6 cm and 8 cm. Find its perimeter.

Answer:

Given :

Diagonal 1 =d1 =8 cm = 8/2 =4

Diagonal 2 = d2 = 6 cm =6/2=3

• By Pythagoras theorem

$AC ^{2} = AB {}^{2} + {}^{2} BC \\ S {}^{2} = D {1}^{2} + D {2}^{2} \\ S { }^{2} = {4}^{2} + {3}^{2} \\ S {}^{2} = 16 + 9 \\ S {}^{2} = \sqrt{25} \\ S = 5$

So, one side = 5 cm

• Perimeter Of Rhombus = 4 × side

= 4 × 5

= 20 cm

Hence, perimeter of such rhombus is 20 cm