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Q.1 \Delta ABC , right angled at A, find BC when:301

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i)  AB=24 \ cm\ \ and\  AC= 7 \ cm\ ,

BC= \sqrt{(24^2+7^2 } =25 \ cm\  

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ii)  AB=28 \ cm\ \ and\  AC=45 \ cm\

 BC=\sqrt{(28^2+45^2 } =53 \ cm\

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iii)  AB=1.6 \ cm\ \ and\  AC=3 \ cm\

 BC=\sqrt{(1.6^2+3^2 } =3.4 \ cm\

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

 BC=\sqrt{(5.6^2+4.2^2 } =7 \ cm\

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Q.2. In  \Delta PQR , right angled at Q, find PQ302

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

  PQ=\sqrt{(29^2-21^2 } =20\ cm\

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ii)   PR=5\ cm\ \ and\  QR=4.8\ cm\

  PQ=\sqrt{(5^2-4.8^2 } =1.4\ cm\

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

  PQ=\sqrt{(3.4^2-3^2 } =1.6\ cm\

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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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Q.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}

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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}

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

Not a right angled triangle

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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}

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

Not a right angled triangle

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Q.4. Find the unknown side of each of the following figures;303

  1. ZY=\sqrt{(32^2+24^2 } =40 \ cm\
  2. LN=\sqrt{(2^2+1.5^2 } =2.5 \ cm\
  3. RQ=\sqrt{(34^2-16^2 } =30 \ cm\
  4. AC=\sqrt{(20^2-12^2 } =16 \ cm\
  5. DF=\sqrt{(0.7^2+2.4^2 } =2.5 \ cm\
  6. KL=\sqrt{(2.5^2+6^2 } =6.5 \ cm\

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Q.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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Q.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 rectangle305

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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Q.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 ABC306

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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Q.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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Q.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 )} =13m

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Q.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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Q.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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Q.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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Q.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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Q.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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Q.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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Q.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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Q.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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Q.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

Answer:

BE^2=AB^2+AE^2 ...i)  

CD^2=AC^2+AD^2 ...ii)  

Adding (1)   &  (2)

BE^2+CD^2=(AB^2+AC^2)  +(AE^2+AD^(2) 

= (2AD) ^2+(2AE) ^2+AE^2+AD^2 

= 5AD^2+5AE^2=5DE^2 

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Q.19. In quadrilateral ABCD, \angle B=\angle D=90^{\circ}. \ Prove \ that \ AB^2-AD^2=CD^2-CB^2  3018

Answer:

AB^2=AC^2-CB^2 ...i)  

AD^2=AC^2-CD^2 ...ii)  

Subtracting ii)  from i)

AB^2-AD^2=CD^2-CB^2 

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