Topics › Geometry › Q156
Rhombus · angle with a diagonal
P6 · Geometry · PSLE Maths
Question
Ali put up a rhombus-shaped paper PQRS in the Maths Corner and drew diagonal PR. Given that ∠PQR = 112°, find ∠QPR, the angle between diagonal PR and side PQ.
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Hint
Look at triangle PQR: in a rhombus, PQ = QR, so it is an isosceles triangle.
Show the answer and step-by-step explanation
Answer: ∠QPR 34 °
- Draw the figure — First draw rhombus PQRS and diagonal PR. The given angle is at the top: ∠PQR = 112°.
- Find isosceles triangle — All sides of a rhombus are equal, so PQ = QR. Therefore, triangle PQR is an isosceles triangle.
- Base angles equal — In isosceles triangle PQR, the base is PR. The base angles ∠QPR and ∠PRQ are equal.
- Find the two base angles — The angles in a triangle add up to 180°. First subtract the vertex angle 112°; the remainder is the sum of the two equal base angles.
- Divide equally — The two base angles are equal, so each base angle is 68° ÷ 2 = 34°. Hence, ∠QPR = 34°. 📌 Key idea: adjacent sides of a rhombus are equal; the diagonal forms an isosceles triangle, so use “remaining angle ÷ 2” to find a base angle.
Method: Adjacent sides of a rhombus are equal. The diagonal forms an isosceles triangle; subtract the vertex angle from 180°, then divide the remaining angle equally between the two base angles.
Practice: 3 similar questions
- In rhombus ABCD, diagonal AC is drawn. Given that ∠ABC = 118°, find ∠BAC.
Answer
31 ° — In a rhombus, AB = BC, so triangle ABC is isosceles. The sum of the two base angles is 180° − 118° = 62°. Each base angle is 62° ÷ 2 = 31°, so ∠BAC = 31°.
- Siti folded a rhombus-shaped bookmark WXYZ along diagonal WY. Given that ∠WXY = 96°, find ∠XWY, the angle between diagonal WY and side WX.
Answer
42 ° — In a rhombus, WX = XY, so triangle WXY is isosceles. The sum of the two equal base angles is 180° − 96° = 84°. Each base angle is 84° ÷ 2 = 42°, so ∠XWY = 42°.
- Raju drew a rhombus PQRS and diagonal PR. Given that the angle between diagonal PR and side PQ is ∠QPR = 27°, find ∠PQR.
Answer
126 ° — In a rhombus, PQ = QR, so triangle PQR is isosceles. Given ∠QPR = 27°, the other base angle ∠PRQ is also 27°. ∠PQR = 180° − 27° − 27° = 126°.
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