Draw rough sketches for the following: (a) In $ \triangle \mathrm{ABC}, \mathrm{BE} $ is a median. (b) In $ \triangle P Q R, P Q $ and PR are altitudes of the triangle. (c) In $ \triangle X Y Z, Y L $ is an altitude in the exterior of the triangle.
Solution :
We have to draw the rough sketches of the following.
(a) In \( \triangle \mathrm{ABC}, \mathrm{BE} \) is a median.
(b) In \( \triangle P Q R, P Q \) and PR are altitudes of the triangle.
(c) In \( \triangle X Y Z, Y L \) is an altitude in the exterior of the triangle.
(a) In \( \triangle \mathrm{ABC}, \mathrm{BE} \) is the median.
(b) $\Delta PQR$ is a right-angled triangle in which PQ and QR are altitudes.
(c) $\Delta XYZ$ is an obtuse-angled triangle YL is an altitude in the exterior of the triangle.
Related Articles Draw rough sketches for the following:(a) In $\Delta P Q R$, PQ, and PR are altitudes of the triangle.(b) In $\Delta XYZ$, YL is an altitude in the exterior of the triangle.
Draw rough sketches for the following:$(a)$ In ∆ABC, BE is a median.$(b)$ In ∆PQR, PQ and PR are altitudes of the triangle.$(c)$ In ∆XYZ, YL is an altitude in the exterior of the triangle.
Construct a triangle \( P Q R \) with side \( Q R=7 \mathrm{~cm}, P Q=6 \mathrm{~cm} \) and \( \angle P Q R=60^{\circ} \). Then construct another triangle whose sides are \( 3 / 5 \) of the corresponding sides of \( \triangle P Q R \).
Study the statements carefully. (P) The line segment joining a vertex of a triangle to the mid point of its opposite side is called a median of the triangle. A triangle has 3 medians.(Q) The perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. A triangle has 3 altitudes. Which of the following options hold?A. Both (P) and (Q) are true. B. Both (P) and (Q) are false.C. (P) is true, (Q) is false.D. (P) is false, (Q) is true.
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\( A, B \) and \( C \) are the midpoints of the sides of \( \Delta X Y Z \). \( P, Q \) and \( R \) are the midpoints of the sides of \( \triangle \mathrm{ABC} \). If \( \mathrm{ABC}=24 \mathrm{~cm}^{2} \), find $XYZ$ and $PQR$.
\( \mathrm{ABC} \) is a triangle. Locate a point in the interior of \( \triangle \mathrm{ABC} \) which is equidistant from all the vertices of \( \triangle \mathrm{ABC} \).
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\( \mathrm{BE} \) and \( \mathrm{CF} \) are two equal altitudes of a triangle \( \mathrm{ABC} \). Using RHS congruence rule, prove that the triangle \( \mathrm{ABC} \) is isosceles.
In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that \( \operatorname{ar}(\Delta \mathrm{PBQ})=a r(\triangle \mathrm{ARC}) \).
In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude. If \( A M=9 \) and \( C M=16 \), find the perimeter of \( \triangle \mathrm{ABC} \).
In figure below, \( \mathrm{PQR} \) is a right triangle right angled at \( \mathrm{Q} \) and \( \mathrm{QS} \perp \mathrm{PR} \). If \( P Q=6 \mathrm{~cm} \) and \( P S=4 \mathrm{~cm} \), find \( Q S, R S \) and \( Q R \)."
Choose the correct answer from the given four options:If in two triangles \( \mathrm{ABC} \) and \( \mathrm{PQR}, \frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}} \), then(A) \( \triangle \mathrm{PQR} \sim \triangle \mathrm{CAB} \)(B) \( \triangle \mathrm{PQR} \sim \triangle \mathrm{ABC} \)(C) \( \triangle \mathrm{CBA} \sim \triangle \mathrm{PQR} \)(D) \( \triangle \mathrm{BCA} \sim \triangle \mathrm{PQR} \)
In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude. If \( \mathrm{BM}=10 \) and \( \mathrm{CM}=5 \), find the perimeter of \( \triangle \mathrm{ABC} \).
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