In Delta ABC,m angle B = 90^{circ}. If AB = 6 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.Here,$AC$ is the hypotenuse and $AB$ is one of the legs.Using th

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.Here,$AC$ is the hypotenuse and $AB$ is one of the legs.Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.Substituting the given values: $10^2 = 6^2 + BC^2$.$100 = 36 + BC^2$.$BC^2 = 100 - 36 = 64$.$BC = \sqrt{64} = 8$.The area of a right-angled triangle is given by the formula: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,base $BC = 8$ and height $AB = 6$.$	ext{Area} = \frac{1}{2} 	imes 8 	imes 6 = 4 	imes 6 = 24$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 6$ and $AC = 10$,find the area of $\Delta ABC$.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

In $\Delta PQR$,$m\angle P = 90^{\circ}$ and $\overline{PM}$ is an altitude. If $PQ = \sqrt{20}$ and $QM = 4$,then $RM = \ldots \ldots .$

$A$ street light bulb is fixed on a pole $6 \, m$ above the level of the street. If a woman of height $1.5 \, m$ casts a shadow of $3 \, m$,find how far she is away from the base of the pole. (in $m$)

In rhombus $ABCD$,$AC = 15$ and $BD = 36$. Find the perimeter of rhombus $ABCD$.

The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module