Class 8: Triangles – Exercise 30D – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: $\Delta ABC$ , right angled at A, find BC when: 301

Answer:

i) $AB=24 \ cm\ \ and\ AC= 7 \ cm\ ,$

$BC= \sqrt{(24^2+7^2 } =25 \ cm\$

ii) $AB=28 \ cm\ \ and\ AC=45 \ cm\$

$BC=\sqrt{(28^2+45^2 } =53 \ cm\$

iii) $AB=1.6 \ cm\ \ and\ AC=3 \ cm\$

$BC=\sqrt{(1.6^2+3^2 } =3.4 \ cm\$

iv) $AB=5.6 \ cm\ \ and\ AC=4.2 \ cm\$

$BC=\sqrt{(5.6^2+4.2^2 } =7 \ cm\$

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Question 2: In $\Delta PQR$ , right angled at Q, find PQ 302

Answer:

i) $PR= 29 \ cm\ \ and\ QR = 21\ cm\$

$PQ=\sqrt{(29^2-21^2 } =20\ cm\$

ii) $PR=5\ cm\ \ and\ QR=4.8\ cm\$

$PQ=\sqrt{(5^2-4.8^2 } =1.4\ cm\$

iii) $PR=3.4\ cm\ \ and\ QR= 3\ cm\$

$PQ=\sqrt{(3.4^2-3^2 } =1.6\ cm\$

iv) $PR=5.3\ cm\ \ and\ QR= 4.5 \ cm\$

$PQ=\sqrt{(5.3^2-4.5^2 } =2.8\ cm\$

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Question 3: The length of the side of some triangles is given below which one of them is right angled? In case of a right angled triangle, find which angle measures $90^{\circ}$ .

Answer:

i) $AB=7 \ cm\ , BC=8 \ cm\ \ and\ AC=15 \ cm\ .$

$BC^2+AC^2=8^2+15^2=289=17^2=AB^2$

$\angle C=90^{\circ}$

ii) $PQ=24 \ cm\ , QR=20 \ cm\ \ and\ PR=32 \ cm\ .$

Not a right angled triangle

iii) $XY=24 \ cm\ , YZ=26 \ cm\ \ and\ ZX=10 \ cm\$

$XY^2+ZX^2=24^2+10^2=676=26^2=YZ$

$\angle X=90^{\circ}$

iv) $LM=1.6\ cm\ , MN=1.4 \ cm\ \ and\ LN=1.8 \ cm\ .$

Not a right angled triangle

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Question 4: Find the unknown side of each of the following figures; 303

(i) $ZY=\sqrt{(32^2+24^2 } =40 \ cm\$

(ii) $LN=\sqrt{(2^2+1.5^2 } =2.5 \ cm\$

(iii) $RQ=\sqrt{(34^2-16^2 } =30 \ cm\$

(iv) $AC=\sqrt{(20^2-12^2 } =16 \ cm\$

(v) $DF=\sqrt{(0.7^2+2.4^2 } =2.5 \ cm\$

(vi) $KL=\sqrt{(2.5^2+6^2 } =6.5 \ cm\$

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Question 5: If the adjoining figure, it is given that: 304

$AD \perp BC, AB=25, AD=15\ cm\ \ and\ AC=17\ cm\$ .Find the length of $i)\ BD, ii)\ DC, iii)\ BC$

Answer:

$BD=\sqrt{(25^2-15^2 } =20 \ cm\$

$DC=\sqrt{(17^2-15^2 } =8 \ cm\$

$BC=BD+DC=28 \ cm\$

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Question 6: If the length of the diagonal of a rectangle is 37 cm. if the length of the shorter side is 12 cm, Find: i) The length of its longer side ii) Perimeter of rectangle iii) Area of rectangle 305

Answer:

$AB=\sqrt{(37^2-12^2 } =35 \ cm\$

Perimeter $= 2\times 35+2\times 12=70+24=94 \ cm\$

Area $=35\times 12=420 \ cm\ ^2$

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Question 7: The Base of an 150 sales triangle in 28 cm long and $AB=AC=50\ cm\$ . Let $AD\perp BC$ find i) Length of AD ii) Area of ABC 306

Answer:

$BD = 14 \ cm\$

$AD=\sqrt{(20^2-14^2 )} =48$

Area of ABC $=1/2\times 28\times 48=672 \ cm\ ^2$

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Question 8: A diagonal of a rhombus in 16 cm long and each of its aides measures 10 cm. Find the length of the other diagonal. 307

Answer:

Diagonal of a rhombus bisect each other and also interest each other at right angle.

$AO=\sqrt{(10^2-8^2 )} =6$

Shorter diagonal $= 12 \ cm$ .

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Question 9: The supporting wire to the top of a vertical pole in 13 cm long and it in fastened to the ground at a state 5 m among from the foot of the pole. How high is the pole? 308

Answer:

Let AB be the pole.

$h=\sqrt{(13^2-5^2 )} =13 \ m$

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Question 10: A ladder 26 m long rust against a vertical wall with its foot 10m against from the wall how high up the wall will the ladder reach. 309

Answer:

Let AB be the height of the ladder.

$h=\sqrt{(26^2-10^2)} =24 \ m$

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Question 11: A 15 meter against a vertical wall 21 reach a window at a height of 12m from the ground. How for in the foot of the ladder? 3010

Answer:

Let the distance of the foot of the ladder from the wall=d

$d=\sqrt{(15^2-12^2) } =9 \ m$

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Question 12: The height of thus to were are 34m and 10m respectively. If the distance between there fact in 32m, find the distance between their tops. 3011

Answer:

Let l be the distance between the tops

$l=\sqrt{(24^2 +32^2) } =40 \ m$

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Question 13: In the adjoining figure, ABC in a triangle in which $\angle B=90^{\circ}$ . If D in the mid point of BC. Prove that $AC^2=AD^2+3CD^2$ 3012

Answer:

$AC^2=AB^2+BC^2$

$=AB^2+(2CD) ^2$

$=(AD^2-BD^2) +4CD^2$

$=AD^2-CD^2+4CD^2$

$AC^2=AD^2+CD^2$

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Question 14: In the adjoining figure, it in given of that $AB=27 \ cm\ , CD=12 \ cm\ , AC=36 \ cm\ \angle BAC=\angle DCA=90^{\circ} \ \\ and\ AM=\ CM\ Find \ i)\ BM^2 \ ii)\ MD^2, \ iii)\ BD^2$ 3013

Answer:

$BM^2=(15+12) ^2+18^2=1053$

$MD^2=18^2+ 12^2=468$

$BD^2=15^2+36^2=1521$

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Question 15: In the following figure, it in being given that: $\angle PQR=90^{\circ}, \angle MNR=90^{\circ} PM=RM, PQ=6 \ cm\ , \\ QR=8 \ cm\ , MN=12 \ cm\$ .

Find the Perimeter of: $\ i)\ \Delta PMR \ ii)\ \Delta MNR \ iii)\ Quadrilateral PQRM$ 3014

Answer:

$PR= \sqrt{(6^2+8^2 )} =10 \ cm\$

$PN=NR= 5 \ cm\$

$MP=MR=\sqrt{(5^2+12^2 )} =13 \ cm\$

Perimeter of $\Delta PMR=13+13+10=36 \ cm\$

Perimeter of $\Delta MNR=12+5+13=30 \ cm\$

Perimeter of $PQRM=13+6+8+13=40 \ cm\$

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Question 16: In the adjoining figure, $\angle PQR=\angle QRS=90^{\circ}. Prove PR^2-PQ^2=QR^2- SR^2$ 3015

Answer:

$PR^2=PQ^2+QR^2 ...i)$

$QS^2=SR^2+QR^2 ...ii)$

Subtract ii) from i) ,

$PR^2-PQ^2=QR^2- SR^2$

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Question 17: In the Adjoining figure, $\angle ABC=90^{\circ}. Prove: AC^2+PQ^2=AQ^2=AQ^2+PC^2$ 3016

Answer:

$AC^2=AB^2+BC^2$

$= (AQ^2-BQ^2) +(PC^2-PB^2)$

$= AQ^2+PC^2-(BQ^2+PB^2)$

$= AQ^2+PC^2-PQ^2$

$\Rightarrow AC^2+PQ^2=AQ^2+PC^2$

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Question 18: In a $\Delta ABC$ , if $D \ and\ E$ are mid point of $AB \ and\ AC$ , respectively and $\ \angle BAC=90^{\circ}, Prove BE^2 +CD^2=5DE^2$ 3017