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Geometry questions for IPMAT, with solutions

10 free Geometry practice questions for IPMAT Indore, Rohtak and JIPMAT, 5 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Geometry on Learn.

1. Question 1Multiple choiceWritten by IPMATwayCircles

   The number of tangents that can be drawn to a circle from a point inside it is

   1. A.$0$
   2. B.infinitely many
   3. C.$2$
   4. D.$1$
   Show the answer and solution

   Answer: A. $0$

   A tangent meets the circle at exactly one point. Any line through a point inside the circle must cross the circle twice (once on each side of the point), so it is a secant, never a tangent. Hence the number of tangents from an interior point is $0$ (from a point on the circle it is $1$, from an outside point $2$).

2. Question 2Type the answerWritten by IPMATwayLines, Angles and Polygons

   Each interior angle of a regular polygon is $4$ times each exterior angle. The number of sides of the polygon is

   Show the answer and solution

   Answer: 10

   At each vertex the interior and exterior angles add to $180^\circ$. Let the exterior angle be $e$; then the interior angle is $4e$, so $4e+e=180^\circ$ and $e=36^\circ$. The exterior angles add to $360^\circ$, so $n=\frac{360^\circ}{36^\circ}=10$ sides.

3. Question 3Type the answerWritten by IPMATwayLines, Angles and Polygons

   A convex polygon has $44$ diagonals. The number of its sides is

   Show the answer and solution

   Answer: 11

   A convex $n$-gon has $\frac{n(n-3)}{2}$ diagonals. So $\frac{n(n-3)}{2}=44$, i.e. $n^2-3n-88=0$. This factors as $(n-11)(n+8)=0$. The number of sides must be positive, so $n=11$.

4. Question 4Type the answerWritten by IPMATwayQuadrilaterals

   In a kite $ABCD$, $AB=AD$ and $CB=CD$. If $\angle ADC=100^\circ$, then $\angle A+\angle C$ (in degrees) is

   Show the answer and solution

   Answer: 160

   In a kite with $AB=AD$ and $CB=CD$, triangles $ABC$ and $ADC$ are congruent (SSS, sharing $AC$). So $\angle ABC=\angle ADC=100^\circ$. The angles of a quadrilateral add to $360^\circ$, so $\angle A+\angle C=360^\circ-100^\circ-100^\circ=160^\circ$.

5. Question 5Multiple choiceAsked in IPMAT Indore 2022Triangles

   In a right-angled triangle $ABC$, the hypotenuse $AC$ is of length $13$ cm. A line drawn connecting the midpoints $D$ and $E$ of sides $AB$ and $AC$ is found to be $6$ cm in length. The length of $BC$ is

   1. A.$12$ cm
   2. B.$5$ cm
   3. C.$2\sqrt{3}$ cm
   4. D.$8$ cm
   Show the answer and solution

   Answer: A. $12$ cm

   $D$ and $E$ are the mid-points of $AB$ and $AC$, so $DE$ is parallel to the third side $BC$ and half of it: $BC=2\cdot 6=12$ cm. (Check: with the right angle at $B$, $AB=\sqrt{169-144}=5$.)

6. Question 6Multiple choiceAsked in IPMAT Indore 2025Circles

   Consider a triangle with side lengths $4$ meters, $6$ meters and $9$ meters. A dog runs around triangle in such a way that the shortest distance of the dog from the triangle is exactly $1$ meter. The total distance covered (in meters) by the dog in one round is:

   1. A.$19+2\pi$
   2. B.$22$
   3. C.$22-2\pi$
   4. D.$22+2\pi$
   Show the answer and solution

   Answer: A. $19+2\pi$

   Along each side the dog runs parallel to it at distance $1$, covering the side's own length. At each corner it swings around on an arc of radius $1$ through the exterior angle; the three exterior angles add to $360^\circ$, so the arcs make one full circle: $2\pi$. Total $19+2\pi$ m.

7. Question 7Multiple choiceAsked in IPMAT Indore 2023Triangles

   In a triangle $ABC$, let $D$ be the mid-point of $BC$, and $AM$ be the altitude on $BC$. If the lengths of $AB$, $BC$ and $CA$ are in the ratio of $2:4:3$, then the ratio of the lengths of $BM$ and $AD$ would be

   1. A.$11:12$
   2. B.$11:4\sqrt{10}$
   3. C.$12:11$
   4. D.$4\sqrt{10}:11$
   Show the answer and solution

   Answer: B. $11:4\sqrt{10}$

   Let $AB=2$, $BC=4$, $CA=3$. From $AB^2-BM^2=AM^2=CA^2-CM^2$ with $CM=4-BM$: $4-BM^2=9-(4-BM)^2\Rightarrow 8BM=11$, $BM=\frac{11}{8}$. Apollonius for the median: $AB^2+AC^2=2AD^2+2BD^2\Rightarrow 13=2AD^2+8\Rightarrow AD=\sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2}$. $BM:AD=\frac{11}{8}:\frac{\sqrt{10}}{2}=11:4\sqrt{10}$.

8. Question 8Multiple choiceAsked in IPMAT Indore 2024Triangles

   Let $ABC$ be a triangle with $AB=AC$ and $D$ be a point on $BC$ such that $\angle BAD=30^\circ$. If $E$ is a point on $AC$ such that $AD=AE$, then $\angle CDE$ equals

   1. A.$15^\circ$
   2. B.$60^\circ$
   3. C.$30^\circ$
   4. D.$10^\circ$
   Show the answer and solution

   Answer: A. $15^\circ$

   With $\angle B=\angle C=\beta$: $\angle BAC=180^\circ-2\beta$ and $\angle DAC=150^\circ-2\beta$. Triangle $ADE$ is isosceles ($AD=AE$), so $\angle ADE=\frac{180^\circ-(150^\circ-2\beta)}{2}=15^\circ+\beta$. $\angle ADC$ is an exterior angle of triangle $ABD$: $\beta+30^\circ$. Hence $\angle CDE=15^\circ$ for every $\beta$ (a classic: $\angle CDE=\frac{1}{2}\angle BAD$).

9. Question 9Multiple choiceAsked in IPMAT Indore 2024Circles

   Let $ABC$ be an equilateral triangle, with each side of length $k$. If a circle is drawn with diameter $AB$, then the area of the portion of the triangle lying inside the circle is

   1. A.$(3\sqrt{3}+\pi)\left(\frac{k^2}{6}\right)$
   2. B.$(3\sqrt{3}-\pi)\left(\frac{k^2}{24}\right)$
   3. C.$(3\sqrt{3}+\pi)\left(\frac{k^2}{24}\right)$
   4. D.$(3\sqrt{3}-\pi)\left(\frac{k^2}{6}\right)$
   Show the answer and solution

   Answer: C. $(3\sqrt{3}+\pi)\left(\frac{k^2}{24}\right)$

   Since $AB$ is a diameter, $\angle AMB=90^\circ$ for $M$ on $AC$, so $M$ is the foot of the perpendicular from $B$ - the mid-point of $AC$ (equilateral). Likewise $N$, the mid-point of $BC$. $MN=\frac{k}{2}$ subtends $60^\circ$ at the centre $O$ (mid-point of $AB$). Area inside $=$ trapezium $ABNM$ $=\frac{3\sqrt{3}}{16}k^2$ plus the circular segment on $MN$: $\frac{60}{360}\pi\frac{k^2}{4}-\frac{\sqrt{3}}{4}\cdot\frac{k^2}{4}$. Total $\frac{3\sqrt{3}k^2}{16}-\frac{\sqrt{3}k^2}{16}+\frac{\pi k^2}{24}=\frac{(3\sqrt{3}+\pi)k^2}{24}$.

10. Question 10Type the answerWritten by IPMATwayLines, Angles and Polygons

    The interior angles of a convex polygon are in arithmetic progression with common difference $5^\circ$, and the smallest angle is $120^\circ$. How many sides does the polygon have?

    Show the answer and solution

    Answer: 9

    The angles are $120,125,\dots,120+5(n-1)$ with sum $\frac n2[240+5(n-1)]$. A polygon's angle sum is $180(n-2)$. So $\frac n2(235+5n)=180n-360\Rightarrow5n^2+235n=360n-720\Rightarrow n^2-25n+144=0\Rightarrow(n-9)(n-16)=0$. Convexity needs the largest angle below $180^\circ$: $120+5(n-1)<180\Rightarrow n<13$. So $n=9$ (largest angle $160^\circ$); $n=16$ would give $195^\circ$, not convex.