MathsCBSE Class 10 · Standard · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 33

From the official CBSE sample question paper for 2023-24, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerTriangles

(a) State and prove Basic Proportionality theorem.

(b) In the given figure $∠ C E F = ∠ C F E$ . F is the midpoint of DC. Prove that $\frac{A B}{B D} = \frac{A E}{F D}$

Figure in the paper: Vertical line AB with A at top and B at bottom. D lies on AB. E lies on AC and C is to the right of D at the same level as the horizontal line DC. F is the midpoint of DC; segment EF is drawn, forming triangle CEF. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

For (a), compare the areas of triangles that share a height, and use the fact that triangles on the same base between the same parallels have equal area.

Hint 2

For (b), you need a line parallel to BE. Try drawing it through D.

Hint 3

With DG ∥ BE, look at △CDG, where F is the midpoint of CD. What does that tell you about E? Then use CF = CE from the equal angles.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle and it meets the other two sides in distinct points, then it divides those two sides in the same ratio.
½ mark
Given: In $△ P QR$ , a line $X Y ∥ QR$ meets $P Q$ at $X$ and $P R$ at $Y$ .
To prove: $\frac{P X}{X Q} = \frac{P Y}{Y R}$
½ mark
Figure and construction: Draw $△ P QR$ with $X$ on $P Q$ and $Y$ on $P R$ , and $X Y ∥ QR$ . Join $Q Y$ and $R X$ . Draw $Y N ⊥ P Q$ (N on PQ) and $X M ⊥ P R$ (M on PR).
½ mark
Proof: Using area $= \frac{1}{2} \times$ base $\times$ height:
$ar (P X Y) = \frac{1}{2} P X \cdot Y N$ and $ar (QX Y) = \frac{1}{2} X Q \cdot Y N$ .
So $\frac{ar ( P X Y )}{ar ( QX Y )} = \frac{P X}{X Q}$ ...(1)
½ mark
Similarly, $ar (P X Y) = \frac{1}{2} P Y \cdot X M$ and $ar (R X Y) = \frac{1}{2} Y R \cdot X M$ .
So $\frac{ar ( P X Y )}{ar ( R X Y )} = \frac{P Y}{Y R}$ ...(2)
½ mark
$△ QX Y$ and $△ R X Y$ stand on the same base $X Y$ and lie between the same parallels $X Y$ and $QR$ . Hence $ar (QX Y) = ar (R X Y)$ ...(3)
From (1), (2) and (3): $\frac{P X}{X Q} = \frac{P Y}{Y R}$ . Hence proved.
½ mark
(b) In the figure, D lies on AB, E lies on AC, F is the midpoint of DC, and B, F, E are collinear (the line BE passes through F).
Construction: Through D, draw $D G ∥ B E$ , meeting AE at G.
In $△ A B E$ , $D G ∥ B E$ , so by BPT $\frac{A D}{D B} = \frac{A G}{GE}$ .
Adding 1 to both sides gives $\frac{A D + D B}{D B} = \frac{A G + GE}{GE}$ , that is, $\frac{A B}{B D} = \frac{A E}{GE}$ ...(i)
½ mark
F is the midpoint of DC, so $C F = F D$ ...(ii)
½ mark
In $△ C D G$ , F is the midpoint of CD and $F E ∥ D G$ (FE is part of BE).
By the converse of the mid-point theorem, E is the midpoint of CG. So $GE = C E$ ...(iii)
½ mark
Given $∠ C E F = ∠ C F E$ , so $C F = C E$ [sides opposite equal angles are equal] ...(iv)
From (ii), (iii) and (iv): $GE = C E = C F = F D$ .
Putting $GE = F D$ in (i): $\frac{A B}{B D} = \frac{A E}{F D}$ . Hence proved.
½ mark

Answer: (a) BPT stated and proved; (b) $\frac{A B}{B D} = \frac{A E}{F D}$ proved (construction DG ∥ BE, then BPT and the converse of the mid-point theorem)

Where the marks are

Correct statement of BPT (½ mark)
Given and To Prove written correctly (½ mark)
Figure with construction (join QY, RX and draw the perpendiculars) (½ mark)
Area ratios with common heights, equal areas of triangles on the same base between the same parallels, and the conclusion (1½ marks)
Constructs DG ∥ BE and applies BPT to get AB/BD = AE/GE (½ mark)
Uses F as the midpoint: CF = FD (½ mark)
Converse of the mid-point theorem in △CDG gives GE = CE (½ mark)
CF = CE from the equal angles, hence GE = FD and the result (½ mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)
Topic: Basic Proportionality Theorem (Thales Theorem)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 33

Hints

Step-by-step solution

Revise the chapter

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