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Tick the correct answer and justify : In Δ ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is :

120°
60°
90°
45°
Explanation:

Given, in ΔABC, AB = 6 $\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm.

We can observe that,

$AB^2 $= 108

$AC^2$ = 144

And, $BC^2$= 36

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

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

$\angle B = 90°$

Hence, the correct answer is (C).

Additional Questions

Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Answer

PQR is a triangle right angled at P and M is a point on QR such that PM $\perp$ QR. Show that $PM^2$ = QM . MR.

Answer

In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that

(i) $AB^2$ = BC . BD

(ii) $AC^2$ = BC . DC

(iii) $AD^2$ = BD . CD

Answer

ABC is an isosceles triangle right angled at C. Prove that $AB^2 = 2AC^2$

Answer

ABC is an isosceles triangle with AC = BC. If $AB^2 = 2 AC^2$, prove that ABC is a right triangle.

Answer

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer

In Fig. 6.54, O is a point in the interior of a triangle ABC, OD $\perp$ BC, OE $\perp$ AC and OF $\perp$ AB. Show that

(i) $OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2$

(ii) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$

Answer

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer
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