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Practical Geometry - Construction of a Right-Angled Triangle (RHS criterion)

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The RHS (Right-angle, Hypotenuse, Side) criterion is a specific condition used to construct a right-angled triangle when the length of its hypotenuse and one of its legs are known. In a right-angled triangle, the side opposite the 90∘90^{\circ} angle is called the hypotenuse, which is always the longest side and appears as a diagonal slope connecting the two legs.

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A rough sketch is essential before starting the actual construction. Draw a triangle with one corner shaped like a square (the 90∘90^{\circ} angle) and label the given side lengths to visualize where the base and hypotenuse will be placed. This helps in identifying which side is the hypotenuse and which is the base.

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The construction begins by drawing the known leg as the base. If we are given ΔABC\Delta ABC right-angled at BB, and side BCBC is given, we draw BCBC as a horizontal line segment. This acts as the foundation of the triangle.

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The right angle is constructed at one of the endpoints of the base using a compass or protractor. Draw a perpendicular ray (often labeled BXBX) pointing upwards from the vertex. Visually, this creates a perfectly vertical 'L' shape relative to the horizontal base.

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The hypotenuse is located using a compass arc. Set the compass width to the given length of the hypotenuse. Place the metal point on the endpoint of the base that is NOT the right-angled vertex. Swing an arc to intersect the perpendicular ray. This intersection point marks the third vertex of the triangle.

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To complete the triangle, join the intersection point on the vertical ray to the endpoint where you placed the compass point. This slanted line segment represents the hypotenuse, completing the three-sided enclosure.

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Verification of the construction can be done using the Pythagoras Theorem. The square of the hypotenuse should be equal to the sum of the squares of the other two sides: a2+b2=c2a^{2} + b^{2} = c^{2}.

📐Formulae

Pythagoras Theorem: a2+b2=c2a^{2} + b^{2} = c^{2} (where cc is the hypotenuse)

Angle Sum Property: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^{\circ}

In any right-angled triangle: Base2+Height2=Hypotenuse2\text{Base}^{2} + \text{Height}^{2} = \text{Hypotenuse}^{2}

💡Examples

Problem 1:

Construct a right-angled triangle ΔPQR\Delta PQR, right-angled at QQ, where QR=3 cmQR = 3\text{ cm} and hypotenuse PR=5 cmPR = 5\text{ cm}.

Solution:

  1. Draw a horizontal line segment QR=3 cmQR = 3\text{ cm} using a ruler.\n2. At point QQ, use a protractor or compass to draw a ray QXQX such that ∠RQX=90∘\angle RQX = 90^{\circ}. This ray should be perpendicular to QRQR.\n3. Set the compass to a radius of 5 cm5\text{ cm}. Place the compass pointer at point RR.\n4. Draw an arc that cuts the ray QXQX at a point. Label this point PP.\n5. Join PP to RR using a ruler. ΔPQR\Delta PQR is the required right-angled triangle.

Explanation:

Since the triangle is right-angled at QQ, QRQR is treated as the base and QXQX is the perpendicular height. The hypotenuse PRPR must connect the far end of the base (RR) to the height (PP). The compass ensures the length PRPR is exactly 5 cm5\text{ cm}.

Problem 2:

Construct ΔLMN\Delta LMN such that ∠M=90∘\angle M = 90^{\circ}, MN=4 cmMN = 4\text{ cm}, and LN=6 cmLN = 6\text{ cm}.

Solution:

  1. Draw the base MN=4 cmMN = 4\text{ cm}.\n2. At vertex MM, construct a 90∘90^{\circ} angle and draw a ray MYMY upwards.\n3. With NN as the center and a radius of 6 cm6\text{ cm}, draw an arc intersecting ray MYMY at point LL.\n4. Join LNLN. The triangle ΔLMN\Delta LMN is constructed.

Explanation:

In this RHS problem, the side MNMN is one leg and LNLN is the hypotenuse (the side opposite the 90∘90^{\circ} angle at MM). We use the compass from NN to find the point LL on the vertical line MYMY.