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Chapter 6: Triangles

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This page provides comprehensive Chapter 6: Triangles - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Triangles. Aligned with the latest rationalised CBSE syllabus.

Standard Level Worksheet

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  1. Question 1: In $\Delta ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE || BC$. If $AD = 4x - 3$, $AE = 8x - 7$, $BD = 3x - 1$ and $CE = 5x - 3$, find the value of $x$.
    Solution:
    Step 1: By BPT, $\frac{AD}{DB} = \frac{AE}{EC}$.
    Step 2: $\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$.
    Step 3: Cross multiply: $(4x-3)(5x-3) = (8x-7)(3x-1)$.
    Step 4: $20x^2 - 12x - 15x + 9 = 24x^2 - 8x - 21x + 7$.
    Step 5: $20x^2 - 27x + 9 = 24x^2 - 29x + 7 \Rightarrow 4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0$.
    Step 6: $(2x+1)(x-1) = 0 \Rightarrow x = 1$ or $x = -0.5$. Since distance cannot be negative ($4x-3$), $x=1$.
    Answer: $x = 1$.
  2. Question 2: $ABCD$ is a trapezium with $AB || DC$. $E$ and $F$ are points on non-parallel sides $AD$ and $BC$ respectively such that $EF$ is parallel to $AB$. Show that $\frac{AE}{ED} = \frac{BF}{FC}$.
    Solution:
    Step 1: Join $AC$ to intersect $EF$ at $G$.
    Step 2: In $\Delta ADC$, $EG || DC$ (since $EF || AB || DC$). By BPT, $\frac{AE}{ED} = \frac{AG}{GC}$.
    Step 3: In $\Delta CAB$, $GF || AB$. By BPT, $\frac{CG}{GA} = \frac{CF}{FB} \Rightarrow \frac{AG}{GC} = \frac{BF}{FC}$.
    Step 4: From steps 2 and 3, $\frac{AE}{ED} = \frac{BF}{FC}$.
  3. Question 3: In the given figure, if $LM || CB$ and $LN || CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
    Solution:
    Step 1: In $\Delta ABC$, $LM || CB$. By BPT, $\frac{AM}{AB} = \frac{AL}{AC}$ (Corollary of BPT).
    Step 2: In $\Delta ADC$, $LN || CD$. By BPT, $\frac{AN}{AD} = \frac{AL}{AC}$.
    Step 3: Comparing both equations, $\frac{AM}{AB} = \frac{AN}{AD}$.
  4. Question 4: A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time, a tower casts a shadow 40 m long on the ground. Determine the height of the tower.
    Solution:
    Step 1: Let height of tower be $h$. Triangles formed by stick/shadow and tower/shadow are similar (Sun's elevation is same).
    Step 2: Ratio of corresponding sides is equal: $\frac{\text{Height of Stick}}{\text{Shadow of Stick}} = \frac{\text{Height of Tower}}{\text{Shadow of Tower}}$.
    Step 3: $\frac{12}{8} = \frac{h}{40} \Rightarrow h = \frac{12 \times 40}{8} = 12 \times 5 = 60$.
    Answer: 60 m.
  5. Question 5: In Figure, $\Delta ODC \sim \Delta OBA$, $\angle BOC = 125^\circ$ and $\angle CDO = 70^\circ$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.
    Solution:
    Step 1: $\angle DOC + \angle BOC = 180^\circ$ (Linear pair). $\angle DOC = 180 - 125 = 55^\circ$.
    Step 2: In $\Delta ODC$, sum of angles is $180^\circ$. $\angle DCO = 180 - (70 + 55) = 180 - 125 = 55^\circ$.
    Step 3: Since $\Delta ODC \sim \Delta OBA$, $\angle OAB = \angle OCD$ (Corresponding angles).
    Step 4: $\angle OAB = 55^\circ$.
    Answer: $55^\circ, 55^\circ, 55^\circ$.
  6. Question 6: $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\Delta ABE \sim \Delta CFB$.
    Solution:
    Step 1: In parallelogram $ABCD$, $\angle A = \angle C$ (Opposite angles).
    Step 2: Since $AE || BC$ (as $AD || BC$), $\angle AEB = \angle CBF$ (Alternate interior angles).
    Step 3: By AA Similarity criterion, $\Delta ABE \sim \Delta CFB$.
  7. Question 7: In $\Delta ABC$, right-angled at $B$, an altitude $BD$ is drawn to the hypotenuse $AC$. If $AB = 6$ cm and $BC = 8$ cm, find the length of $BD$.
    Solution:
    Step 1: In $\Delta ABC$, $AC = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10$ cm.
    Step 2: We know $\Delta ADB \sim \Delta ABC$. So $\frac{BD}{BC} = \frac{AB}{AC}$.
    Step 3: $BD = \frac{AB \times BC}{AC} = \frac{6 \times 8}{10} = \frac{48}{10} = 4.8$ cm.
    Answer: 4.8 cm.
  8. Question 8: $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
    Solution:
    Step 1: In $\Delta ABC$ and $\Delta DAC$:
    $\angle BAC = \angle ADC$ (Given)
    $\angle C = \angle C$ (Common)
    Step 2: By AA Similarity, $\Delta ABC \sim \Delta DAC$.
    Step 3: Corresponding sides are proportional: $\frac{CA}{CD} = \frac{CB}{CA}$.
    Step 4: Cross multiplying: $CA^2 = CB \cdot CD$.
  9. Question 9: Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\Delta PQR$. Show that $\Delta ABC \sim \Delta PQR$.
    Solution:
    Step 1: Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$. Since $AD, PM$ are medians, $BC = 2BD$ and $QR = 2QM$.
    Step 2: $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
    Step 3: By SSS similarity, $\Delta ABD \sim \Delta PQM$. This implies $\angle B = \angle Q$.
    Step 4: In $\Delta ABC$ and $\Delta PQR$: $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
    Step 5: By SAS similarity, $\Delta ABC \sim \Delta PQR$.
  10. Question 10: In Figure, $DE || OQ$ and $DF || OR$. Show that $EF || QR$.
    Solution:
    Step 1: In $\Delta POQ$, $DE || OQ$. By BPT, $\frac{PE}{EQ} = \frac{PD}{DO}$.
    Step 2: In $\Delta POR$, $DF || OR$. By BPT, $\frac{PF}{FR} = \frac{PD}{DO}$.
    Step 3: From (1) and (2), $\frac{PE}{EQ} = \frac{PF}{FR}$.
    Step 4: In $\Delta PQR$, since sides $PQ$ and $PR$ are divided in the same ratio, by Converse of BPT, $EF || QR$.
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