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Answers
FOR 1st FIGURE.....
QSR is a right angled ∆
So, by Pythagoras therom
→ h² = p² + b²
→ 30² = (QR)² + 18²
→ (QR)² = 30² - 18²
→ (QR)² = 576
So, QR = 24
SO, PQRS is a rectangle
Hence perimeter of rectangle
= 2( l + b )
= 2(24 + 18)
= 84 cm __________________(Ans.)
FOR 2nd FIGURE.....
It's an isosceles ∆
FD = DE (given)
EF = 5.5 cm
DE = FD = 2EF = 2 x 5.5 = 11 cm
So, perimeter of ∆
= FD + DE + EF
= 11 + 11 + 5.5
= 27.5 cm _________________(Ans.)
FOR 3rd FIGURE....
∆ BCD is right angled triangle
CD = 12 cm, BC = 13 cm & BD = ?
Again, by Pythagoras therom
→ h² = p² + b²
→ (BC)² = (CD)² + (BD)²
→ 13² = 12² + (BD)²
→ (BD)² = 169 - 144
→ (BD)² = 25
SO, BD = 5
And, AB = AD + BD
→ 14 = AD + 5
→ AD = 14 - 5
So, AD = 9
NOW, In ∆ ACD
→ h² = p² = b²
→ (AC)² = (CD)² + (AD)²
→ (AC)² = 12² + 9²
→ (AC)² = 144 + 81
SO, AC = 15
Hence,
Perimeter of ∆ ABC is
= AC + BC + AB
= 15 + 13 + 14
= 42 cm ___________________(Ans.)
Answer:
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