In Delta ABC,AB = 17,BC = 15,and AC = 8. Then | vedclass

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

Question

(A) Given sides of $\Delta ABC$ are $AB = 17$,$BC = 15$,and $AC = 8$.First,check if the triangle is a right-angled triangle using the converse of the

Vedclass · Accepted Answer

$8.5$

Vedclass · Answer

(A) Given sides of $\Delta ABC$ are $AB = 17$,$BC = 15$,and $AC = 8$.First,check if the triangle is a right-angled triangle using the converse of the Pythagorean theorem.$BC^2 + AC^2 = 15^2 + 8^2 = 225 + 64 = 289$.$AB^2 = 17^2 = 289$.Since $BC^2 + AC^2 = AB^2$,$\Delta ABC$ is a right-angled triangle with the hypotenuse $AB = 17$.The longest side of a right-angled triangle is its hypotenuse.According to the property of a right-angled triangle,the length of the median drawn to the hypotenuse is half the length of the hypotenuse.Length of median = $\frac{1}{2} 	imes AB = \frac{1}{2} 	imes 17 = 8.5$.

In $\Delta ABC$,$AB = 17$,$BC = 15$,and $AC = 8$. Then,the length of the median on the longest side of the triangle is $\ldots$.

Similar Questions

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{PM}$ is a median. Prove that $PR^{2} = PM^{2} + 3RM^{2}$.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 3x - 19$,$MB = x - 3$,$MC = x - 5$,and $MD = 3$,find the value of $x$.

$\square ABCD$ is a rectangle. If $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$,find the length of diagonal $\overline{AC}$.

In $\Delta ABC$,$AB > AC$ and $D$ is the midpoint of $\overline{BC}$. $\overline{AM} \perp \overline{BC}$ and $M \in \overline{BC}$. Prove that $AB^{2} - AC^{2} = 2 \cdot BC \cdot DM$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude. If $PD = 9RD$,prove that $PQ = 3QR$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module