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Class 9 Triangles

Overview

This Class 9 Triangles HOTS worksheet challenges you to prove results about triangle inequalities, medians, angle bisectors and congruence. Build each argument from the stated conditions and established triangle properties.

HOTS (High Order Thinking Skills)

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  1. Question 1: In a triangle, if the bisector of an angle is perpendicular to the opposite side, prove that the triangle is isosceles.
    Solution: Let $AD$ be the bisector of $\angle A$ perpendicular to $BC$. In $\triangle ABD$ and $\triangle ACD$: $\angle BAD = \angle CAD$, $AD = AD$, $\angle ADB = \angle ADC = 90^\circ$. By ASA, $\triangle ABD \cong \triangle ACD$. Thus $AB = AC$.
  2. Question 2: Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.
    Solution: Extend median $AD$ to $E$ such that $AD = DE$. Join $EC$. $\triangle ABD \cong \triangle ECD$ (SAS). So $AB = EC$. In $\triangle AEC$, $AC + EC > AE \Rightarrow AC + AB > 2AD$.
  3. Question 3: $O$ is a point in the interior of a triangle $ABC$. Show that $OA + OB + OC > \frac{1}{2} (AB + BC + CA)$.
    Solution: In $\triangle OAB$, $OA + OB > AB$. In $\triangle OBC$, $OB + OC > BC$. In $\triangle OCA$, $OC + OA > CA$. Adding all: $2(OA + OB + OC) > AB + BC + CA$. Divide by 2.
  4. Question 4: Prove that the perimeter of a triangle is greater than the sum of its three medians.
    Solution: We know $AB + AC > 2AD$, $AB + BC > 2BE$, $BC + AC > 2CF$. Adding these: $2(AB + BC + AC) > 2(AD + BE + CF) \Rightarrow AB + BC + AC > AD + BE + CF$.
  5. Question 5: Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of triangle $PQR$. Show that $\triangle ABC \cong \triangle PQR$.
    Solution: $BC = QR \Rightarrow BM = QN$. In $\triangle ABM$ and $\triangle PQN$: $AB=PQ, AM=PN, BM=QN$. So $\triangle ABM \cong \triangle PQN$ (SSS). Thus $\angle B = \angle Q$. Now in $\triangle ABC$ and $\triangle PQR$: $AB=PQ, \angle B=\angle Q, BC=QR$. So $\triangle ABC \cong \triangle PQR$ (SAS).
  6. Question 6: If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.
    Solution: Let $AD$ be bisector of $\angle A$ and $BD=CD$. Extend $AD$ to $E$ such that $AD=DE$. Join $CE$. $\triangle ABD \cong \triangle ECD$. So $AB=CE$ and $\angle BAD = \angle CED$. Since $\angle BAD = \angle CAD$, $\angle CAD = \angle CED$. So $AC=CE$. Thus $AB=AC$.
  7. Question 7: $ABC$ is a triangle in which $\angle B = 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$. Prove that $\angle BAC = 72^\circ$.
    Solution: Construct line $AE$ such that $\angle CAE = \angle C$. Then $AE=EC$. Use exterior angle properties and congruence to show relations leading to $x + 2x + 2x = 180$ or similar logic. (Detailed proof involves construction).
  8. Question 8: $S$ is any point on side $QR$ of a triangle $PQR$. Show that $PQ + QR + RP > 2PS$.
    Solution: In $\triangle PQS$, $PQ + QS > PS$. In $\triangle PRS$, $PR + SR > PS$. Adding both: $PQ + PR + (QS + SR) > 2PS \Rightarrow PQ + PR + QR > 2PS$.
  9. Question 9: Prove that the difference between any two sides of a triangle is less than the third side.
    Solution: In $\triangle ABC$, $AB + BC > AC \Rightarrow AB > AC - BC$. Similarly for other sides. Thus difference is always less than the third side.
  10. Question 10: In triangle $ABC$, the bisector $AD$ of $\angle A$ is perpendicular to side $BC$. Show that $AB = AC$ and triangle $ABC$ is isosceles.
    Solution: In $\triangle ABD$ and $\triangle ACD$: $\angle BAD = \angle CAD$ (Bisector), $AD = AD$ (Common), $\angle ADB = \angle ADC = 90^\circ$ (Perpendicular). By ASA, $\triangle ABD \cong \triangle ACD$. Hence $AB = AC$.
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