In Delta ABC,angle B = 90^{circ},AB = 14 , cm | 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 $\angle B = 90^{\circ}$.Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.Substitute the

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(D) Given that $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.Substitute the given values: $50^2 = 14^2 + BC^2$.$2500 = 196 + BC^2$.$BC^2 = 2500 - 196 = 2304$.$BC = \sqrt{2304} = 48 \, cm$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.$	ext{Area} = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 14 	imes 48$.$	ext{Area} = 7 	imes 48 = 336 \, cm^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

Similar Questions

In the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.

In $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.

In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.

In $\Delta XYZ$,points $A, B, C, D, E, F,$ and $G$ lie on the side $YZ$ such that $YA = AB = BC = CD = DE = EF = FG = GZ$. Prove that $ar(XBE) = \frac{3}{8} ar(XYZ)$.

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module