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Identify the shapes and their properties.

Mathematics

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2 parts
a

5 cm5 \text{ cm}

b

513 cm5\sqrt{13} \text{ cm}

Step-by-step solution

8 steps

Here's an analysis of the figures provided:

Part a· 3 steps

Triangle ABC

This is a right-angled triangle with the right angle at A. We are given the hypotenuse BC=13 cmBC = 13 \text{ cm} and one leg AC=12 cmAC = 12 \text{ cm}. We can find the length of the other leg, ABAB, using the Pythagorean theorem.

  1. Apply the Pythagorean theorem.

    AB2+AC2=BC2AB^2 + AC^2 = BC^2

  2. Substitute the given values.

    AB2+(12 cm)2=(13 cm)2AB^2 + (12 \text{ cm})^2 = (13 \text{ cm})^2 AB2+144 cm2=169 cm2AB^2 + 144 \text{ cm}^2 = 169 \text{ cm}^2

  3. Solve for ABAB.

    AB2=169 cm2−144 cm2AB^2 = 169 \text{ cm}^2 - 144 \text{ cm}^2 AB2=25 cm2AB^2 = 25 \text{ cm}^2 AB=25 cm2AB = \sqrt{25 \text{ cm}^2} AB=5 cmAB = \boxed{5 \text{ cm}}

Part b

513 cm5\sqrt{13} \text{ cm}

Triangle DEF

This is a right-angled triangle with the right angle at E. We are given the lengths of the two legs, DE=15 cmDE = 15 \text{ cm} and EF=10 cmEF = 10 \text{ cm}. We can find the length of the hypotenuse, DFDF, using the Pythagorean theorem.

  1. Apply the Pythagorean theorem.

    DE2+EF2=DF2DE^2 + EF^2 = DF^2

  2. Substitute the given values.

    (15 cm)2+(10 cm)2=DF2(15 \text{ cm})^2 + (10 \text{ cm})^2 = DF^2 225 cm2+100 cm2=DF2225 \text{ cm}^2 + 100 \text{ cm}^2 = DF^2

  3. Solve for DFDF.

    DF2=325 cm2DF^2 = 325 \text{ cm}^2 DF=325 cm2DF = \sqrt{325 \text{ cm}^2} DF=25×13 cm2DF = \sqrt{25 \times 13 \text{ cm}^2} DF=513 cmDF = \boxed{5\sqrt{13} \text{ cm}}

Part c· 1 step

Quadrilateral ABCD

  1. Identify the properties of the quadrilateral.

    The figure shows a quadrilateral with all four angles marked as right angles (90∘90^\circ). The adjacent sides have different lengths (AB=10 cmAB = 10 \text{ cm} and BC=15 cmBC = 15 \text{ cm}).
    This shape is a rectangle.

Part d· 1 step

Quadrilateral PQRS

  1. Identify the properties of the quadrilateral.

    The figure shows a quadrilateral with all four angles marked as right angles (90∘90^\circ). The adjacent sides have equal lengths (PQ=10 cmPQ = 10 \text{ cm} and QR=10 cmQR = 10 \text{ cm}).
    This shape is a square.

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