In Delta ABC,m angle C = 90^circ,AB = 12.5 an | vedclass

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

Question

(C) Given that $\Delta ABC$ is a right-angled triangle with $m \angle C = 90^\circ$.According to the Pythagorean theorem,in a right-angled triangle,th

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(C) Given that $\Delta ABC$ is a right-angled triangle with $m \angle C = 90^\circ$.According to the Pythagorean theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$AB$ is the hypotenuse,and $AC$ and $BC$ are the other two sides.Therefore,$AB^2 = AC^2 + BC^2$.Substituting the given values: $(12.5)^2 = AC^2 + (12)^2$.$156.25 = AC^2 + 144$.$AC^2 = 156.25 - 144$.$AC^2 = 12.25$.Taking the square root of both sides,$AC = \sqrt{12.25} = 3.5$.

In $\Delta ABC$,$m \angle C = 90^\circ$,$AB = 12.5$ and $BC = 12$. Find $AC$.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A : m \angle C = 1 : 2$. If $BC = 4$,find $AC$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AB = 10.8$,$MB = 4.5$ and $AC = 4.8$,find $AN$.

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle P$ and $\angle B \cong \angle Q$. If $\frac{AB}{PQ} = \frac{4}{5}$ and $AC = 20$,find $PR$.

If $\Delta XYZ \sim \Delta DEF$ for the correspondence $XYZ \leftrightarrow EFD$.
If $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$,then in $\Delta DEF$,$\ldots \ldots$ is a right angle.

In $\Delta ABC$,$m \angle A = 90^{\circ}$,$AB = 8$ and $AC = 15$. Then,$BC = \ldots$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module