Question

answer please guys please answer correct is not free point time please help me only answer correct is my request​

Answer 1

2) Given :

⬤ PQ = 10 cm .

⬤ PR = 24 cm .

To Find :

⬤ Value of QR .

Formula Used :

$\bf[\:{{(Hypotenuse)}^{2}={(Base)}^{2}+{(Perpendicular)}^{2}}\:]$

Solution :

By Pythagoras Theorem :-

In ∆ PQR :

(H)² = (B)² + (P)²

(QR)² = (PQ)² + (PR)²

(QR)² = (10)² + (24)²

(QR)² = (10×10) + (24×24)

(QR)² = 100 + 576

(QR)² = 676

QR = √676

QR = 26

Therefore , The Value of QR is 26 cm .

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3) Given :

⬤ A tree is broken at a height of 5 m from the Ground .

⬤ Its top touches the ground at a distance of 12 m from Base of the tree .

To Find :

⬤ Original Height of Tree .

Formula Used :

$\bf[\:{{(Hypotenuse)}^{2}={(Base)}^{2}+{(Perpendicular)}^{2}}\:]$

Solution :

By Pythagoras Theorem :-

In ∆ ABC :

(AC)² = (BC)² + (AB)²

(AC)² = (12)² + (5)²

(AC)² = (12×12) + (5×5)

(AC)² = 144 + 25

(AC)² = 169

AC = √169

AC = 13

Therefore , The Value of AC is 13 .

Now :

Original Height of Tree = 13 + 5

= 18

Therefore , The Original Height of Tree is 18 m .