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Triangles

Short Answer Type

481. In figure, AD π CD and BC π CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.

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482. In ∆ABC, D is the mid point of BC. The perpendiculars from D to AB and AC are equal. Prove that ∆ABC is isosceles.

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483. In figure, ∆ABD and ∆BCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.

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Long Answer Type

484. In ∆’s ABC and PQR, AB = PQ, AC = PR and altitude AM and altitude PN are equal. Show that ∆ABC ≅ ∆PQR.

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485. In figure, the perpendicular AD, BE and CF drawn from the vertices A, B and C respectively of ∆ABC are equal. Prove that the triangle is an equilateral triangle.

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486. ABCD is a parallelogram. If the two diagonals are equal, find the measure of ∠ABC.

In ∆ADB and ∆BCA,

In ∆ADB and ∆BCA,
AD = BC| Opposite sides of a parallelogram are

AD = BC
| Opposite sides of a parallelogram are equal
AB = BA    | Common
DB = CA    | Given
∴ ∆ADB ≅ ∆BCA
| SSS congruence rule
∴ DAB = ∠CBA    ...(1) | CPCT
But AD || BC
| Opposite sides of parallelogram ABCD and a transversal AB intersects then
∴ ∠DAB + ∠CBA = 180°
| Sum of consecutive interior angles on the same side of a transversal is 180°
⇒ ∠CBA + ∠CBA = 180° | From (1)
⇒ 2∠CB A = 180°
⇒ ∠CBA = 90°
⇒ ∠ABC = 90°

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Short Answer Type

487. AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see figure). Show that the line PQ is the perpendicular bisector of AB.

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488. P is a point equidistant from two lines l and m intersecting at point A (see figure). Show that the line AP bisects the angle between them.

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489.

In the figure, RS = QT and QS = RT. Prove that PQ = PR.

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490. In the given figure, D is the mid-point of base BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ∠B = ∠C.

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